CF445E：DZY Loves Colors（线段树区间更新）^

E. DZY Loves Colors

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

1. Paint all the units with numbers between l and r (both inclusive) with color x.
2. Ask the sum of colorfulness of the units between l and r (both inclusive).

Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

input

3 31 1 2 41 2 3 52 1 3

output

8

input

3 41 1 3 42 1 12 2 22 3 3

output

321

input

10 61 1 5 31 2 7 91 10 10 111 3 8 121 1 10 32 1 10

output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.

题意：N个点初始贡献为0，颜色为i，M个操作：1是将区间内的颜色变成val，且每个点的贡献为abs(x-val)。2是求出区间内每个数的历史贡献之和。

思路：线段树区间更新，区间内的颜色不一样怎么办呢？col[]数组储存区间内颜色是否一样，一样才更新lazy之类的，否则继续往下递归。

# include <bits/stdc++.h># define lson l,m,id<<1# define rson m+1,r,id<<1|1using namespace std;typedef long long LL;const int maxn = 1e5+30;LL col[maxn<<2], sum[maxn<<2], lazy[maxn<<2];void build(int l, int r, int id){    if(l==r)    {        col[id] = l;        return;    }    int m = l+r>>1;    build(lson);    build(rson);    sum[id] = sum[id<<1] + sum[id<<1|1];    if(col[id<<1] && col[id<<1]==col[id<<1|1]) col[id] = col[id<<1];    else col[id] = 0;}void pushdown(int rt, int dis){    if(lazy[rt])    {        lazy[rt<<1] += lazy[rt];        lazy[rt<<1|1] += lazy[rt];        col[rt<<1] = col[rt<<1|1] = col[rt];        sum[rt << 1] += lazy[rt]*(dis-(dis>>1));        sum[rt<<1|1] += lazy[rt]*(dis>>1);        lazy[rt] = 0;    }}void update(int L, int R, int val, int l, int r, int id){    if(L<=l && R>=r && col[id])    {        lazy[id] += abs(col[id]-val);        sum[id] += (LL)llabs(val-col[id])*(r-l+1);        col[id] = val;        return;    }    pushdown(id, r-l+1);    int m = l+r>>1;    if(L <= m) update(L, R, val, lson);    if(R > m) update(L, R, val, rson);    sum[id] = sum[id<<1] + sum[id<<1|1];    if(col[id<<1] && col[id<<1]==col[id<<1|1]) col[id] = col[id<<1];    else col[id] = 0;}LL query(int L, int R, int l, int r, int id){    if(L<=l && R>=r)        return sum[id];    pushdown(id, r-l+1);    LL ans = 0;    int m = l+r>>1;    if(L <= m) ans += query(L, R, lson);    if(R > m) ans += query(L, R, rson);    return ans;}int main(){    int n, m;    scanf("%d%d",&n,&m);    build(1, n, 1);    while(m--)    {        int op, l, r, x;        scanf("%d%d%d",&op,&l,&r);        if(op == 1)        {            scanf("%d",&x);            update(l,r,x,1,n,1);        }        else            printf("%I64d\n",query(l,r,1,n,1));    }    return 0;}