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;
const int N = 100010;
int n, m, a[N];
long long c[N];
struct block {
  int color, l, r;
  long long sum;
  long long add;
} b[1000];
int in(int x, int B) { return (x - 1) / B + 1; }
int main() {
  scanf("%d%d", &n, &m);
  int B = sqrt(n + 0.5);
  int num = n / B;
  for (int i = 1; i <= num + 1; ++i) {
    b[i].color = -1;
    b[i].sum = b[i].add = 0;
    b[i].l = (i - 1) * B + 1;
    b[i].r = i * B;
  }
  for (int i = 1; i <= n; ++i) a[i] = i;
  memset(c, 0, sizeof(c));
  int t, l, r, x;
  while (m--) {
    scanf("%d%d%d", &t, &l, &r);
    int lx = in(l, B), rx = in(r, B);
    if (t == 1) {
      scanf("%d", &x);
      if (lx == rx) {
        int ix = lx;
        if (b[ix].color != -1) {
          for (int i = b[ix].l; i <= b[ix].r; ++i) a[i] = b[ix].color;
        }
        b[ix].color = -1;
        for (int i = l; i <= r; ++i) {
          c[i] += abs(x - a[i]);
          b[ix].sum += abs(x - a[i]);
          a[i] = x;
        }
      } else {
        int st = lx + 1, ed = rx - 1;
        if (b[lx].color != -1) {
          for (int i = b[lx].l; i <= b[lx].r; ++i) a[i] = b[lx].color;
        }
        b[lx].color = -1;
        for (int i = l; i <= b[lx].r; ++i) {
          c[i] += abs(x - a[i]);
          b[lx].sum += abs(x - a[i]);
          a[i] = x;
        }
        if (b[rx].color != -1) {
          for (int i = b[rx].l; i <= b[rx].r; ++i) a[i] = b[rx].color;
        }
        b[rx].color = -1;
        for (int i = b[rx].l; i <= r; ++i) {
          c[i] += abs(x - a[i]);
          b[rx].sum += abs(x - a[i]);
          a[i] = x;
        }
        for (int i = st; i <= ed; ++i) {
          if (b[i].color == -1) {
            for (int j = b[i].l; j <= b[i].r; ++j) {
              b[i].sum += abs(a[j] - x);
              c[j] += abs(a[j] - x);
            }
            b[i].color = x;
          } else {
            b[i].add += abs(x - b[i].color);
            b[i].sum += (long long)abs(x - b[i].color) * (long long)B;
            b[i].color = x;
          }
        }
      }
    } else {
      long long ans = 0;
      if (lx == rx) {
        int ix = lx;
        for (int i = l; i <= r; ++i) ans += (c[i] + b[ix].add);
      } else {
        int st = lx + 1, ed = rx - 1;
        for (int i = st; i <= ed; ++i) ans += b[i].sum;
        for (int i = l; i <= b[lx].r; ++i) ans += b[lx].add + c[i];
        for (int i = b[rx].l; i <= r; ++i) ans += b[rx].add + c[i];
      }
      printf("%I64d\n", ans);
    }
  }
  return 0;
}
