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 = 1e6 + 10;
const int SQ = 450;
long long C[N], C_baze[SQ], point[N], point_kole_baze[SQ], point_baze[SQ], n, m;
void upd_kamel(int ind, int l, int r, int x) {
  long long st = ind * SQ, en = min(n, st + SQ - 1);
  if (C_baze[ind] == -1) {
    for (int i = st; i <= en; i++) {
      point[i] += abs(C[i] - x);
      point_kole_baze[ind] += abs(C[i] - x);
      C[i] = x;
    }
    C_baze[ind] = x;
  } else {
    point_kole_baze[ind] += abs(C_baze[ind] - x) * (en - st + 1);
    point_baze[ind] += abs(C_baze[ind] - x);
    C_baze[ind] = x;
  }
}
void upd_nesfe(int ind, int l, int r, int x) {
  long long st = ind * SQ, en = min(n, st + SQ - 1);
  if (C_baze[ind] != -1) {
    for (int i = st; i <= en; i++) {
      C[i] = C_baze[ind];
      point[i] += point_baze[ind];
    }
    C_baze[ind] = -1;
    point_baze[ind] = 0;
  }
  for (int i = l; i <= r; i++) {
    point[i] += abs(C[i] - x);
    point_kole_baze[ind] += abs(C[i] - x);
    C[i] = x;
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i <= n; i++) C[i] = i;
  for (int i = 0; i < SQ; i++) C_baze[i] = -1;
  while (m--) {
    int type;
    cin >> type;
    if (type == 1) {
      int l, r, x;
      cin >> l >> r >> x;
      int st = l / SQ, en = r / SQ;
      if (st == en)
        upd_nesfe(st, l, r, x);
      else {
        upd_nesfe(st, l, (st + 1) * SQ - 1, x);
        for (long long i = st + 1; i < en; i++)
          upd_kamel(i, i * SQ, (i + 1) * SQ - 1, x);
        upd_nesfe(en, en * SQ, r, x);
      }
    } else {
      int l, r;
      cin >> l >> r;
      long long ans = 0;
      for (int i = l; i <= r; i++) {
        if ((i % SQ) == 0 && (i + SQ - 1) <= r)
          ans += point_kole_baze[i / SQ], i += (SQ - 1);
        else
          ans += (point[i] + point_baze[i / SQ]);
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
