codeforce 742 D. Arpa's weak amphitheater and Mehrdad's valuable Hoses (分组背包+并查集）)

有关系的并查集分在一组~01背包~每组选一个或全选~更新最优解即可~~

Just to remind, girls in Arpa's land are really nice.

Mehrdad wants to invite some Hoses to the palace for a dancing party. Each Hos has some weightw_i and some beautyb_i. Also each Hos may have some friends. Hoses are divided in some friendship groups. Two Hosesx and y are in the same friendship group if and only if there is a sequence of Hosesa₁, a₂, ..., a_k such thata_i anda_i + 1 are friends for each1 ≤ i < k, and a₁ = x and a_k = y.

Arpa allowed to use the amphitheater of palace to Mehrdad for this party. Arpa's amphitheater can hold at mostw weight on it.

Mehrdad is so greedy that he wants to invite some Hoses such that sum of their weights is not greater thanw and sum of their beauties is as large as possible. Along with that, from each friendship group he can either invite all Hoses, or no more than one. Otherwise, some Hoses will be hurt. Find for Mehrdad the maximum possible total beauty of Hoses he can invite so that no one gets hurt and the total weight doesn't exceedw.

Input

The first line contains integers n, m and w (1  ≤  n  ≤  1000,,1 ≤ w ≤ 1000) — the number of Hoses, the number of pair of friends and the maximum total weight of those who are invited.

The second line contains n integers w₁, w₂, ..., w_n (1 ≤ w_i ≤ 1000) — the weights of the Hoses.

The third line contains n integers b₁, b₂, ..., b_n (1 ≤ b_i ≤ 10⁶) — the beauties of the Hoses.

The next m lines contain pairs of friends, thei-th of them contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n,x_i ≠ y_i), meaning that Hosesx_i andy_i are friends. Note that friendship is bidirectional. All pairs(x_i, y_i) are distinct.

Output

Print the maximum possible total beauty of Hoses Mehrdad can invite so that no one gets hurt and the total weight doesn't exceedw.

Examples

Input

3 1 53 2 52 4 21 2

Output

6

Input

4 2 112 4 6 66 4 2 11 22 3

Output

7

Note

In the first sample there are two friendship groups: Hoses {1, 2} and Hos {3}. The best way is to choose all of Hoses in the first group, sum of their weights is equal to5 and sum of their beauty is 6.

In the second sample there are two friendship groups: Hoses {1, 2, 3} and Hos {4}. Mehrdad can't invite all the Hoses from the first group because their total weight is12 > 11, thus the best way is to choose the first Hos from the first group and the only one from the second group. The total weight will be8, and the total beauty will be 7.

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>using namespace std;const int N = 1e3 + 100;struct node{int v,w;}a[N];vector<node>vec[N];int fa[N],dp[N];int find(int x){int r=x;while(fa[r]!=r) r=fa[r];int i=x,j;while(i!=r) {j=fa[i];fa[i]=r;i=j;}return r;}int main(){int n,v,w,i,j,m,t1,t2,k,V,W;cin>>n>>m>>w;for(i=1;i<=n;i++) cin>>a[i].v;  //体积 for(i=1;i<=n;i++) cin>>a[i].w;  //价值 for(i=1;i<=n;i++) fa[i]=i;while(m--) {cin>>t1>>t2;if(t1>t2) swap(t1,t2);int fx=find(t1),fy=find(t2);if(fx!=fy) fa[fy]=fx;}for(i=1;i<=n;i++) {vec[find(i)].push_back(a[i]);}node tmp,t;for(i=1;i<=n;i++) {if(find(i)==i) {tmp.v=tmp.w=0;for(j=0;j<vec[i].size();j++) {t=vec[i][j];tmp.v+=t.v,tmp.w+=t.w;}vec[find(i)].push_back(tmp);}}for(i=1;i<=n;i++) {      if(find(i)==i) for(j=w;j>=0;j--) {for(int z=0;z<vec[find(i)].size();z++) {tmp=vec[i][z];if(tmp.v<=j) dp[j]=max(dp[j],dp[j-tmp.v]+tmp.w);}}}cout<<dp[w]<<endl;return 0;}