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 = (1 << 18) + 10;
int n, m;
long long tree[N], add[N];
int same[N];
int col[N];
void build(int x, int l, int r) {
  tree[x] = same[x] = col[x] = add[x] = 0;
  if (l == r) {
    same[x] = 1;
    col[x] = l;
    return;
  }
  int s = (l + r) / 2;
  build(x + x, l, s);
  build(x + x + 1, s + 1, r);
}
void push(int x, int l, int r) {
  tree[x] += (long long)(r - l + 1) * add[x];
  if (l != r) {
    add[x + x] += add[x];
    add[x + x + 1] += add[x];
    if (same[x]) {
      same[x + x] = same[x + x + 1] = 1;
      col[x + x] = col[x + x + 1] = col[x];
    }
  }
  add[x] = 0;
}
long long upd(int x, int l, int r, int tl, int tr, int y) {
  push(x, tl, tr);
  if (tr < l || tl > r) return tree[x];
  if (l <= tl && tr <= r) {
    if (same[x]) {
      add[x] += abs(col[x] - y);
      col[x] = y;
      push(x, tl, tr);
      return tree[x];
    } else
      same[x] = 1, col[x] = y;
  } else if (same[x] && col[x] != y)
    col[x] = same[x] = 0;
  int tm = (tl + tr) / 2;
  return tree[x] =
             upd(x + x, l, r, tl, tm, y) + upd(x + x + 1, l, r, tm + 1, tr, y);
}
long long get(int x, int l, int r, int tl, int tr) {
  push(x, tl, tr);
  if (tr < l || tl > r) return 0;
  if (l <= tl && tr <= r) return tree[x];
  int tm = (tl + tr) / 2;
  return get(x + x, l, r, tl, tm) + get(x + x + 1, l, r, tm + 1, tr);
}
int main() {
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  int t, l, r, x;
  for (int i = 0; i < m; i++) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d%d", &l, &r, &x);
      upd(1, l, r, 1, n, x);
    } else {
      scanf("%d%d", &l, &r);
      printf("%I64d\n", get(1, l, r, 1, n));
    }
  }
  return 0;
}
