#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
int block, l[N], r[N];
int belong[N];
int num;
long long a[N];
int n, m;
long long b[N];
long long ls[N], lz[N];
long long extra[N];
void build() {
  block = sqrt(n);
  num = n / block;
  if (n % block) num++;
  for (int i = 1; i <= num; i++) {
    l[i] = (i - 1) * block + 1;
    r[i] = i * block;
  }
  r[num] = n;
  for (int i = 1; i <= n; i++) {
    belong[i] = (i - 1) / block + 1;
  }
}
int main() {
  scanf("%d%d", &n, &m);
  build();
  for (int i = 1; i <= n; i++) a[i] = i;
  while (m--) {
    int cmd;
    int L, R;
    long long w;
    scanf("%d", &cmd);
    if (cmd == 1) {
      scanf("%d%d%I64d", &L, &R, &w);
      for (int i = L; i <= R;) {
        int k = belong[i];
        int y = min(R, r[k]);
        if (i == l[k] && y == r[k]) {
          if (!ls[k]) {
            for (int o = l[k]; o <= r[k]; o++) {
              b[o] += (long long)fabs(a[o] - w);
              extra[k] += (long long)fabs(a[o] - w);
              a[o] = w;
            }
            ls[k] = w;
          } else {
            extra[k] += (r[k] - l[k] + 1) * ((long long)fabs(ls[k] - w));
            lz[k] += (long long)fabs(ls[k] - w);
            ls[k] = w;
          }
          i = r[k] + 1;
        } else {
          if (ls[k]) {
            for (int o = l[k]; o <= r[k]; o++) {
              a[o] = ls[k];
            }
            ls[k] = 0;
          }
          b[i] += (long long)fabs(a[i] - w);
          extra[k] += (long long)fabs(a[i] - w);
          a[i] = w;
          i++;
        }
      }
    } else {
      scanf("%d%d", &L, &R);
      long long sum = 0;
      for (int i = L; i <= R;) {
        int k;
        k = belong[i];
        int y = min(R, r[k]);
        if (i == l[k] && y == r[k]) {
          sum += extra[k];
          i = r[k] + 1;
        } else {
          sum += b[i] + lz[k];
          i++;
        }
      }
      printf("%I64d\n", sum);
    }
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
int block, l[N], r[N];
int belong[N];
int num;
long long a[N];
int n, m;
long long b[N];
long long ls[N], lz[N];
long long extra[N];
void build() {
  block = sqrt(n);
  num = n / block;
  if (n % block) num++;
  for (int i = 1; i <= num; i++) {
    l[i] = (i - 1) * block + 1;
    r[i] = i * block;
  }
  r[num] = n;
  for (int i = 1; i <= n; i++) {
    belong[i] = (i - 1) / block + 1;
  }
}
int main() {
  scanf("%d%d", &n, &m);
  build();
  for (int i = 1; i <= n; i++) a[i] = i;
  while (m--) {
    int cmd;
    int L, R;
    long long w;
    scanf("%d", &cmd);
    if (cmd == 1) {
      scanf("%d%d%I64d", &L, &R, &w);
      for (int i = L; i <= R;) {
        int k = belong[i];
        int y = min(R, r[k]);
        if (i == l[k] && y == r[k]) {
          if (!ls[k]) {
            for (int o = l[k]; o <= r[k]; o++) {
              b[o] += (long long)fabs(a[o] - w);
              extra[k] += (long long)fabs(a[o] - w);
              a[o] = w;
            }
            ls[k] = w;
          } else {
            extra[k] += (r[k] - l[k] + 1) * ((long long)fabs(ls[k] - w));
            lz[k] += (long long)fabs(ls[k] - w);
            ls[k] = w;
          }
          i = r[k] + 1;
        } else {
          if (ls[k]) {
            for (int o = l[k]; o <= r[k]; o++) {
              a[o] = ls[k];
            }
            ls[k] = 0;
          }
          b[i] += (long long)fabs(a[i] - w);
          extra[k] += (long long)fabs(a[i] - w);
          a[i] = w;
          i++;
        }
      }
    } else {
      scanf("%d%d", &L, &R);
      long long sum = 0;
      for (int i = L; i <= R;) {
        int k;
        k = belong[i];
        int y = min(R, r[k]);
        if (i == l[k] && y == r[k]) {
          sum += extra[k];
          i = r[k] + 1;
        } else {
          sum += b[i] + lz[k];
          i++;
        }
      }
      printf("%I64d\n", sum);
    }
  }
}

```