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 nm = 100005;
int n, k, m, t;
long long col[nm << 2];
long long ness[nm << 2], more[nm << 2];
void Init(int i, int l, int r) {
  if (l != r) {
    Init((i << 1), l, ((l + r) >> 1));
    Init((i << 1) + 1, ((l + r) >> 1) + 1, r);
    return;
  }
  col[i] = l;
}
void Up(int i) {
  while (i >= 1) {
    if (col[(i << 1)] == col[(i << 1) + 1])
      col[i] = col[(i << 1)];
    else
      col[i] = 0;
    ness[i] = ness[(i << 1)] + ness[(i << 1) + 1];
    i >>= 1;
  }
}
void Downvalue(int i, int l, int r) {
  col[(i << 1)] = col[(i << 1) + 1] = col[i];
  ness[(i << 1)] += more[i] * (((l + r) >> 1) - l + 1);
  ness[(i << 1) + 1] += more[i] * (r - ((l + r) >> 1));
  more[(i << 1)] += more[i];
  more[(i << 1) + 1] += more[i];
  more[i] = col[i] = 0;
}
void Update(int i, int l, int r, int x, int y, int w) {
  if (r < x || l > y) return;
  if (l >= x && r <= y && col[i]) {
    more[i] += fabs(w - col[i]);
    ness[i] += (r - l + 1) * fabs(w - col[i]);
    col[i] = w;
    Up(i >> 1);
    return;
  }
  if (l == r) return;
  if (col[i]) Downvalue(i, l, r);
  if (((l + r) >> 1) >= x) Update((i << 1), l, ((l + r) >> 1), x, y, w);
  if (((l + r) >> 1) + 1 <= y)
    Update((i << 1) + 1, ((l + r) >> 1) + 1, r, x, y, w);
}
long long Get(int i, int l, int r, int x, int y) {
  if (r < x || l > y) return 0;
  if (l >= x && r <= y) {
    return ness[i];
  }
  if (col[i] != 0) Downvalue(i, l, r);
  return Get((i << 1), l, ((l + r) >> 1), x, y) +
         Get((i << 1) + 1, ((l + r) >> 1) + 1, r, x, y);
}
int main() {
  int i, x, y, z, w;
  scanf("%d%d", &n, &m);
  Init(1, 1, n);
  for (i = 1; i <= m; i++) {
    scanf("%d", &z);
    if (z == 1) {
      scanf("%d%d%d", &x, &y, &w);
      Update(1, 1, n, x, y, w);
    } else {
      scanf("%d%d", &x, &y);
      printf("%I64d\n", Get(1, 1, n, x, y));
    }
  }
}
