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.
#include <bits/stdc++.h>
using namespace std;
long long n, m, t, l, r, tam, a[100010], x, bloco[1000], clf[100010],
    bclf[1000], bclft[1000];
int main() {
  scanf("%lld %lld", &n, &m);
  for (long long i = 0; i < n; i++) a[i] = i + 1;
  tam = sqrt(n);
  for (long long i = 0; i < m; i++) {
    scanf("%lld %lld %lld", &t, &l, &r);
    l--;
    if (t == 1) {
      scanf("%lld", &x);
      while (l < r) {
        if (l % tam || (r - l < tam && r < n)) {
          if (bloco[l / tam]) {
            for (long long j = l - (l % tam); j < l - (l % tam) + tam && j < n;
                 j++)
              a[j] = bloco[l / tam];
            bloco[l / tam] = 0;
          }
          clf[l] += abs(x - a[l]);
          bclft[l / tam] += abs(x - a[l]);
          a[l] = x;
          l++;
        } else {
          if (bloco[l / tam]) {
            bclf[l / tam] += abs(x - bloco[l / tam]);
            bclft[l / tam] += min(tam, r - l) * abs(x - bloco[l / tam]);
          } else {
            for (long long j = l; j < r && j < l + tam; j++) {
              bclft[l / tam] += abs(x - a[j]);
              clf[j] += abs(x - a[j]);
            }
          }
          bloco[l / tam] = x;
          l += tam;
        }
      }
    } else {
      long long resp = 0;
      while (l < r) {
        if (l % tam || (r - l < tam && r < n)) {
          while ((l % tam || r - l < tam) && l < r) {
            resp += bclf[l / tam] + clf[l];
            l++;
          }
        } else {
          resp += bclft[l / tam];
          l += tam;
        }
      }
      printf("%lld\n", resp);
    }
  }
  return 0;
}
