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;
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
void file() {}
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
const int N = 1e5 + 9;
long long arr[N], sum[N], sumall[N], color[N], val[N];
int main() {
  file();
  fast();
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) arr[i] = i + 1;
  int sq = sqrt(n);
  while (q--) {
    int t, l, r;
    cin >> t >> l >> r;
    l--, r--;
    if (t == 1) {
      int x;
      cin >> x;
      for (int j = l; j <= r; j++) {
        if (j % sq == 0 && j + sq - 1 <= r) {
          if (color[j / sq])
            sum[j / sq] += abs(x - color[j / sq]),
                sumall[j / sq] += abs(x - color[j / sq]) * sq;
          else
            for (int tmp = j; tmp <= j + sq - 1; tmp++)
              val[tmp] += abs(arr[tmp] - x),
                  sumall[j / sq] += abs(arr[tmp] - x);
          color[j / sq] = x;
          j += sq - 1;
        } else {
          if (color[j / sq]) {
            int tmp = (j / sq) * sq;
            while (tmp / sq == j / sq) arr[tmp++] = color[j / sq];
            color[j / sq] = 0;
          }
          val[j] += abs(arr[j] - x), sumall[j / sq] += abs(arr[j] - x),
              arr[j] = x;
        }
      }
    } else {
      long long ans = 0;
      for (int j = l; j <= r; j++) {
        if (j % sq == 0 && j + sq - 1 <= r)
          ans += sumall[j / sq], j += sq - 1;
        else
          ans += val[j] + sum[j / sq];
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
