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>
#pragma GCC optimize("Ofast")
using namespace std;
long long n, sn, m, i, i2, i3, tmp, l, r, x, a[101001], b[101001], ta[401],
    sumb[401], tb[401], ans, th, tl, tr;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m, sn = sqrt(n);
  for (i = 0; i < n; i++) a[i] = i + 1;
  while (m--) {
    cin >> tmp >> l >> r, l--, r--;
    if (tmp == 1) {
      cin >> x;
      for (i = l; i <= r;) {
        th = i / sn, tl = th * sn, tr = min(tl + sn, n);
        if (i == tl && r >= tr - 1) {
          if (ta[th])
            tb[th] += abs(ta[th] - x), sumb[th] += abs(ta[th] - x) * (tr - tl);
          else
            for (i2 = tl; i2 < tr; i2++)
              b[i2] += abs(a[i2] - x), sumb[th] += abs(a[i2] - x);
          ta[th] = x, i = tr;
        } else {
          if (ta[th]) {
            for (i2 = tl; i2 < tr; i2++) a[i2] = ta[th];
            ta[th] = 0;
          }
          b[i] += abs(a[i] - x), sumb[th] += abs(a[i] - x), a[i] = x, i++;
        }
      }
    } else {
      ans = 0;
      for (i = l; i <= r;) {
        tl = i / sn * sn, tr = min(tl + sn, n), th = i / sn;
        if (i == tl && r >= tr - 1)
          ans += sumb[th], i = tr;
        else
          ans += tb[th] + b[i], i++;
      }
      cout << ans << '\n';
    }
  }
}
