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 maxn = 1e5 + 500;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, unit, a[maxn];
long long extra[maxn], lz[maxn], ls[maxn], b[maxn];
int main() {
  n = read(), m = read();
  unit = sqrt(n);
  for (int i = 0; i < n; ++i) a[i] = i + 1;
  while (m--) {
    int op = read(), l = read(), r = read(), x;
    --l;
    --r;
    if (op == 1) {
      x = read();
      for (int i = l; i <= r;) {
        int idx = i / unit;
        int st = idx * unit;
        int ed = min(n, (idx + 1) * unit);
        if (i == st && r >= ed - 1) {
          if (ls[idx]) {
            extra[idx] += 1LL * abs(x - ls[idx]) * (ed - st);
            lz[idx] += abs(x - ls[idx]);
            ls[idx] = x;
          } else {
            for (int j = st; j < ed; ++j) {
              b[j] += abs(a[j] - x);
              extra[idx] += abs(a[j] - x);
              a[j] = x;
            }
            ls[idx] = x;
          }
          i = ed;
        } else {
          if (ls[idx]) {
            for (int j = st; j < ed; ++j) a[j] = ls[idx];
            ls[idx] = 0;
          }
          extra[idx] += abs(a[i] - x);
          b[i] += abs(a[i] - x);
          a[i] = x;
          ++i;
        }
      }
    } else {
      long long res = 0;
      for (int i = l; i <= r;) {
        int idx = i / unit;
        int st = idx * unit;
        int ed = min(n, (idx + 1) * unit);
        if (i == st && r >= ed - 1) {
          res += extra[idx];
          i = ed;
        } else {
          res += b[i] + lz[idx];
          ++i;
        }
      }
      printf("%lld\n", res);
    }
  }
  return 0;
}
