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 tag_A[350], sum_B[350], tag_B[350], A[100000], B[100000];
int main() {
  long long i, j, n, q, sqn, k, ql, qr, l, r, x, id, re;
  cin >> n >> q;
  sqn = sqrt(n);
  for (i = 0; i < n; i++) A[i] = i + 1;
  while (q--) {
    cin >> k >> ql >> qr;
    ql--, qr--;
    if (k == 1) {
      cin >> x;
      for (i = ql; i <= qr;) {
        id = i / sqn, l = id * sqn, r = min(n, l + sqn);
        if (i == l && qr >= r - 1) {
          if (tag_A[id])
            sum_B[id] += abs(tag_A[id] - x) * (r - l),
                tag_B[id] += abs(tag_A[id] - x);
          else
            for (j = l; j < r; j++)
              B[j] += abs(A[j] - x), sum_B[id] += abs(A[j] - x);
          tag_A[id] = x, i = r;
        } else {
          if (tag_A[id]) {
            for (j = l; j < r; j++) A[j] = tag_A[id];
            tag_A[id] = 0;
          }
          sum_B[id] += abs(A[i] - x), B[i] += abs(A[i] - x), A[i] = x;
          i++;
        }
      }
    } else {
      re = 0;
      for (i = ql; i <= qr;) {
        id = i / sqn, l = id * sqn, r = min(n, l + sqn);
        if (i == l && qr >= r - 1)
          re += sum_B[id], i = r;
        else
          re += B[i] + tag_B[id], i++;
      }
      cout << re << "\n";
    }
  }
}
