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 warning(disable : 4996)
using namespace std;
int n, m;
long long col[411111], val[411111], fl[411111], ql, qr, qd;
void bld(int x, int l, int r) {
  if (l != r)
    bld((x << 1), l, ((l + r) >> 1)),
        bld(((x << 1) | 1), ((l + r) >> 1) + 1, r);
  else
    col[x] = l;
}
void push_clo(int x, int l, int r) {
  col[(x << 1)] = col[((x << 1) | 1)] = col[x];
  col[x] = 0;
}
void push_val(int x, int l, int r) {
  fl[(x << 1)] += fl[x], fl[((x << 1) | 1)] += fl[x];
  val[(x << 1)] += fl[x] * (((l + r) >> 1) - l + 1),
      val[((x << 1) | 1)] += fl[x] * (r - ((l + r) >> 1));
  fl[x] = 0;
}
void maintain(int x, int l, int r) {
  col[x] = col[(x << 1)] == col[((x << 1) | 1)] ? col[(x << 1)] : 0;
  val[x] = val[(x << 1)] + val[((x << 1) | 1)];
}
void upd(int x, int l, int r) {
  if (ql <= l && qr >= r) {
    if (col[x]) {
      val[x] += abs(qd - col[x]) * (r - l + 1);
      fl[x] += abs(qd - col[x]);
      col[x] = qd;
    } else {
      upd((x << 1), l, ((l + r) >> 1)),
          upd(((x << 1) | 1), ((l + r) >> 1) + 1, r);
      maintain(x, l, r);
    }
  } else {
    if (fl[x]) push_val(x, l, r);
    if (col[x]) push_clo(x, l, r);
    if (ql <= ((l + r) >> 1)) upd((x << 1), l, ((l + r) >> 1));
    if (qr > ((l + r) >> 1)) upd(((x << 1) | 1), ((l + r) >> 1) + 1, r);
    maintain(x, l, r);
  }
}
long long ans;
void que(int x, int l, int r) {
  if (ql <= l && qr >= r)
    ans += val[x];
  else {
    if (fl[x]) push_val(x, l, r);
    if (col[x]) push_clo(x, l, r);
    if (ql <= ((l + r) >> 1)) que((x << 1), l, ((l + r) >> 1));
    if (qr > ((l + r) >> 1)) que(((x << 1) | 1), ((l + r) >> 1) + 1, r);
  }
}
int main() {
  int t, x, y, z, ca = 1;
  scanf("%d%d", &(n), &(m));
  int ty = 0;
  bld(1, 1, n);
  while (m--) {
    scanf("%d", &(ty));
    if (ty == 1) {
      scanf("%d%d%d", &(ql), &(qr), &(qd));
      upd(1, 1, n);
    } else {
      scanf("%d%d", &(ql), &(qr));
      ans = 0;
      que(1, 1, n);
      printf("%I64d\n", (ans));
    }
  }
  return 0;
}
