Codeforces Round #395 (Div. 1) E. Timofey and our friends animals(lct动态维护mst)

After his birthday party, Timofey went to his favorite tree alley in a park. He wants to feed there his favorite birds — crows.

It's widely known that each tree is occupied by a single crow family. The trees in the alley form a row and are numbered from 1 to n. Some families are friends to each other. For some reasons, two families can be friends only if they live not too far from each other, more precisely, there is no more than k - 1 trees between any pair of friend families. Formally, the family on the u-th tree and the family on thev-th tree can be friends only if |u - v| ≤ k holds.

One of the friendship features is that if some family learns that Timofey is feeding crows somewhere, it notifies about this all friend families. Thus, after Timofey starts to feed crows under some tree, all the families that are friends to the family living on this tree, as well as their friends and so on, fly to the feeding place. Of course, the family living on the tree also comes to the feeding place.

Today Timofey came to the alley and noticed that all the families that live on trees with numbers strictly less than l or strictly greater thanr have flown away. Thus, it is not possible to pass the information about feeding through them. Moreover, there is no need to feed them. Help Timofey to learn what is the minimum number of trees under which he has to feed crows so that all the families that have remained will get the information about feeding. You are given several situations, described by integers l and r, you need to calculate the answer for all of them.

Input

The first line contains integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 5), where n is the number of trees, and k is the maximum possible distance between friend families.

The next line contains single integer m (0 ≤ m ≤ n·k) — the number of pair of friend families.

Each of the next m lines contains two integers u and v (1 ≤ u, v ≤ 105), that means that the families on trees u and v are friends. It is guaranteed that u ≠ v and |u - v| ≤ k. All the given pairs are distinct.

The next line contains single integer q (1 ≤ q ≤ 105) — the number of situations you need to calculate the answer in.

Each of the next q lines contains two integers l and r (1 ≤ l ≤ r ≤ 105), that means that in this situation families that have flown away lived on such trees x, so that either x < l or x > r.

Output

Print q lines. Line i should contain single integer — the answer in the i-th situation.

Example

input

5 331 32 34 551 11 22 31 31 5

output

12112

Note

In the first example the following family pairs are friends: (1, 3), (2, 3) and (4, 5).

• In the first situation only the first family has remained, so the answer is 1.
• In the second situation the first two families have remained, and they aren't friends, so the answer is 2.
• In the third situation the families 2 and 3 are friends, so it is enough to feed any of them, the answer is 1.
• In the fourth situation we can feed the first family, then the third family will get the information from the first family, and the second family will get the information from the third. The answer is 1.

• In the fifth situation we can feed the first and the fifth families, so the answer is 2.

题意：给n个点，m条边，然后q个询问，每次询问[l,r]区间中所有点带上其两两点之间的边组成的图的连通分量个数。

分析：思路很神奇，我们把询问离线按照左端排序，然后从大到小枚举左端点，对于每个左端点 l 我们把所有E(u,v) (u < v && u >=  l)的边扔进来，求出这些边集的最小生成树(边权为max(u,v)),那么对于一个询问(l,r)，此时我们只要求出当前最小生成树中权值小于等于r的边的个数就可以了(设其为x,这样答案就是区间长度 - x),在左端点移动时就会产生一个动态维护mst的问题，每次新加进来来一个边E(x,y)我们先判断一下此时(x,y)的连通性，如果已经连通还需要删掉(x,y)中的最大边,再把当前的边加进来，这整个过程要用动态树维护(中间TLE了好几次，又犯了上次cut时直接删点的错误。。).

#include<iostream>#include<string>#include<algorithm>#include<cstdlib>#include<cstdio>#include<set>#include<map>#include<vector>#include<cstring>#include<stack>#include<queue>#define INF 2147483640#define eps 1e-9#define MAXN 0x3f#define N 100005using namespace std;int n,m,k,x,y,q,s[6*N],ch[6*N][2],f[N],Ans[N],value[6*N],Max[6*N],Max_num[6*N],fa[6*N];bool lazy[6*N];struct thing{int x,y,num,val;}ask[N],edg[6*N];vector<int> G_ask[N],G_edg[N];inline bool isroot(int x){return ch[fa[x]][0] != x && ch[fa[x]][1] != x;}void push_up(int x){Max[0] = 0;Max[x] = value[x],Max_num[x] = x;if(Max[x] < Max[ch[x][0]]){Max[x] = Max[ch[x][0]];Max_num[x] = Max_num[ch[x][0]];}if(Max[x] < Max[ch[x][1]]){Max[x] = Max[ch[x][1]];Max_num[x] = Max_num[ch[x][1]];}}void rotate(int x){int y = fa[x],z = fa[y];int d = ch[y][0] == x ? 0 : 1;if(!isroot(y)){if(ch[z][0] == y) ch[z][0] = x;else ch[z][1] = x;}fa[y] = x,fa[x] = z,fa[ch[x][d^1]] = y;ch[y][d] = ch[x][d^1],ch[x][d^1] = y;push_up(y),push_up(x);}inline void push_down(int x){if(!lazy[x]) return;int ls = ch[x][0],rs = ch[x][1];lazy[x]^=1,lazy[ls]^=1;lazy[rs]^=1;swap(ch[ls][0],ch[ls][1]);swap(ch[rs][0],ch[rs][1]);}void splay(int x){int tot = 0;s[++tot] = x;for(int i = x;!isroot(i);i = fa[i]) s[++tot] = fa[i];for(;tot;tot--) push_down(s[tot]);while(!isroot(x)){int y = fa[x],z = fa[y];if(!isroot(y)){if((ch[z][0] == y) ^ (ch[y][0] == x)) rotate(x);else rotate(y);}rotate(x);}}void access(int x){int t = 0;while(x){splay(x);ch[x][1] = t;          //parent path 边中儿子一定是子链root，父亲是子链root实际父亲 push_up(x);t = x,x = fa[x];}}int findroot(int x){splay(x);while(ch[x][0]) x = ch[x][0];return x;}void makeroot(int x)        //x变成根 {access(x),splay(x);swap(ch[x][0],ch[x][1]);lazy[x]^=1; }void link(int x,int y){makeroot(x);fa[x] = y;}void cut(int x,int y)                          //一次一条边 {makeroot(x),access(y),splay(y),ch[y][0] = fa[x] = 0;push_up(y);}bool islink(int x,int y){makeroot(x);access(y);return findroot(y) == x;}int lowbit(int x){return x & (-x);}void Insert(int x,int k){while(x <= n){f[x] += k;x += lowbit(x);}}int Find_sum(int x){int ans = 0;while(x){ans += f[x];x -= lowbit(x);}return ans;}int main(){memset(Max,0,sizeof(Max));memset(value,0,sizeof(value));scanf("%d%d",&n,&k);scanf("%d",&m);for(int i = 1;i <= m;i++){scanf("%d%d",&edg[i].x,&edg[i].y);if(edg[i].x > edg[i].y) swap(edg[i].x,edg[i].y); edg[i].val = edg[i].y;Max[i + n] = value[i + n] = edg[i].val;Max_num[i + n] = i + n;G_edg[edg[i].x].push_back(i);}scanf("%d",&q);for(int i = 1;i <= q;i++){scanf("%d%d",&ask[i].x,&ask[i].y);G_ask[ask[i].x].push_back(i);}for(int i = n;i;i--){for(int j = 0;j < G_edg[i].size();j++){int e = G_edg[i][j];int x = edg[e].x,y = edg[e].y,val = edg[e].val;if(islink(x,y)){if(Max[y] > val){int u = edg[Max_num[y] - n].x,v = edg[Max_num[y] - n].y,t = Max_num[y];Insert(Max[y],-1);Insert(val,1);cut(u,t);                          //删过Max_num[y]就会变 cut(t,v);link(x,n + e);link(n + e,y);}}else {Insert(val,1);link(x,n + e);link(n + e,y);}}for(int j = 0;j < G_ask[i].size();j++) {int v = G_ask[i][j];Ans[v] = ask[v].y - ask[v].x + 1 - Find_sum(ask[v].y);}}for(int i = 1;i <= q;i++) printf("%d\n",Ans[i]);}