Codeforces Round #254 (Div. 1) -- C. DZY Loves Colors（分块）

题意：

给你n 个数，有q 个操作，操作1，使得L到R的区间的a全变成x，b 加上|x-a|。 操作2查询L到R上的b 总和。

思路：

分块乱搞就好了。

我们用flag[i]表示i 这个分块是否是一样的。

当前分块如果是完整的话，如果是一样的话，直接更新到答案里，不是的话就暴力一遍。

如果不完整的话，在讨论是不是一样的。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int maxn = 1e5+10;

LL a[maxn],b[maxn];
int belong[maxn];
int L[maxn], R[maxn];
int block;
int num;
LL sum[maxn];

int n, q;

LL flag[maxn], flag2[maxn];


void init(){
    for (int i = 1; i <= n; ++i) a[i] = i;
    int num = sqrt(n);
    block = n / num;
    if (n % block) ++num;

    for (int i = 1; i <= num ;++i){
        L[i] = (i-1)*block + 1;
        R[i] = i*block;
    }
    R[n] = num;
    for (int i = 1 ; i <= n; ++i){
        belong[i] = (i-1)/block+1;
    }


}

LL aaa(LL x){
    if (x < 0) return -x;
    return x;
}
void update(int l,int r,int x){

    for (int i = l; i <= r;){
        int id = belong[i];
        if (i == L[id] && r >= R[id]){
            if (flag[id]){
                sum[id] += (LL)(R[id]-L[id]+1) * aaa(x-flag[id]);
                flag2[id] += aaa(x-flag[id]);
                flag[id] = x;
            }
            else {
                flag[id] = x;
                for (int j = L[id]; j <= R[id]; ++j){
                    sum[id] += aaa(a[j] - x);
                    b[j] += aaa(a[j] - x);
                    a[j] = x;
                }
            }
            i = R[id]+1;
        }
        else {
            if (flag[id]){
                for (int j = L[id]; j <= R[id]; ++j)a[j] = flag[id];
                flag[id] = 0;
            }
            sum[id] += aaa(a[i]-x);
            b[i] += aaa(a[i]-x);
            a[i] = x;
            ++i;
        }
    }

}

LL query(int l,int r){
    LL ans = 0LL;
    for (int i = l; i <= r; ){
        int id = belong[i];
        if (i == L[id] && r >= R[id]){
            ans += sum[id];
            i = R[id]+1;
        }
        else {
            ans += b[i] + flag2[id];
            ++i;
        }


    }
    return ans;
}
int main(){
    scanf("%d %d",&n, &q);
    init();
    int op;
    while(q--){
        scanf("%d",&op);
        if (op == 1){
            int l,r,v;
            scanf("%d %d %d",&l, &r, &v);
            update(l,r,v);
        }
        else {
            int l,r;
            scanf("%d %d",&l, &r);
            printf("%lld\n",query(l,r));
        }
    }
    return 0;
}

C. 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 3
1 1 2 4
1 2 3 5
2 1 3

Output

8

Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3

Output

3
2
1

Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 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.