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;
struct maca {
  long long int b, v, g;
} mg[100005];
struct kuca {
  long long int p, k, s, ta, tb;
} gr[1000];
int main() {
  ios::sync_with_stdio(false);
  long long int i, j, k, m, n, h, koren, res, grupa, ug = 0, pp, l, d, x;
  cin >> n >> m;
  koren = sqrt(n);
  j = 1;
  grupa = 1;
  ug = 1;
  gr[1].p = 1;
  for (i = 1; i <= n; i++, j++) {
    mg[i].b = i;
    mg[i].v = 0;
    if (j <= koren)
      mg[i].g = grupa;
    else {
      gr[ug].k = i - 1;
      grupa++;
      j = 1;
      mg[i].g = grupa;
      ug++;
      gr[ug].p = i;
    }
  }
  gr[ug].k = n;
  for (h = 1; h <= m; h++) {
    cin >> pp >> l >> d;
    if (pp == 1) {
      cin >> x;
      for (i = 1; i <= ug; i++) {
        if (gr[i].p > d)
          continue;
        else if (gr[i].k < l)
          continue;
        else if (gr[i].p >= l && gr[i].k <= d) {
          if (gr[i].ta) {
            gr[i].tb += abs(gr[i].ta - x);
            gr[i].s += abs(gr[i].ta - x) * (gr[i].k - gr[i].p + 1);
          } else {
            for (j = gr[i].p; j <= gr[i].k; j++) {
              mg[j].v += abs(mg[j].b - x);
              gr[i].s += abs(mg[j].b - x);
            }
          }
          gr[i].ta = x;
        } else {
          if (gr[i].ta) {
            for (j = gr[i].p; j <= gr[i].k; j++) mg[j].b = gr[i].ta;
            gr[i].ta = 0;
          }
          for (j = gr[i].p; j <= gr[i].k; j++) {
            if (j >= l && j <= d) {
              mg[j].v += abs(mg[j].b - x);
              gr[i].s += abs(mg[j].b - x);
              mg[j].b = x;
            }
          }
        }
      }
    } else {
      res = 0;
      for (i = 1; i <= ug; i++) {
        if (gr[i].p > d)
          continue;
        else if (gr[i].k < l)
          continue;
        else if (gr[i].p >= l && gr[i].k <= d)
          res = res + gr[i].s;
        else {
          for (j = gr[i].p; j <= gr[i].k; j++) {
            if (j >= l && j <= d) res = res + mg[j].v + gr[i].tb;
          }
        }
      }
      cout << res << endl;
    }
  }
  return 0;
}
