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;
struct SegmentTree {
  long long l, r, col, sum, lazy;
} tree[4444444];
int n, m;
void build(int x, int l, int r) {
  tree[x].l = l;
  tree[x].r = r;
  if (l == r) {
    tree[x].col = l;
    return;
  }
  int mid = (l + r) / 2;
  build(x << 1, l, mid);
  build(x << 1 | 1, mid + 1, r);
}
void spread(int x, int d) {
  if (tree[x].col) {
    tree[x].lazy += abs(tree[x].col - d);
    tree[x].sum += abs(tree[x].col - d) * (tree[x].r - tree[x].l + 1);
    return;
  }
  spread(x << 1, d);
  spread(x << 1 | 1, d);
  tree[x].sum = tree[x << 1].sum + tree[x << 1 | 1].sum +
                tree[x].lazy * (tree[x].r - tree[x].l + 1);
}
void update(int x, int l, int r, int d) {
  if (l <= tree[x].l && r >= tree[x].r) {
    spread(x, d);
    tree[x].col = d;
    return;
  }
  if (tree[x].col) {
    tree[x << 1].col = tree[x << 1 | 1].col = tree[x].col;
    tree[x].col = 0;
  }
  int mid = (tree[x].l + tree[x].r) / 2;
  if (l <= mid) update(x << 1, l, r, d);
  if (r > mid) update(x << 1 | 1, l, r, d);
  tree[x].sum = tree[x << 1].sum + tree[x << 1 | 1].sum +
                tree[x].lazy * (tree[x].r - tree[x].l + 1);
}
long long ask(int x, int l, int r) {
  if (l <= tree[x].l && r >= tree[x].r) return tree[x].sum;
  long long ans = 0;
  int mid = (tree[x].l + tree[x].r) / 2;
  if (l <= mid) ans += ask(x << 1, l, min(mid, r));
  if (r > mid) ans += ask(x << 1 | 1, max(mid + 1, l), r);
  return ans + tree[x].lazy * (r - l + 1);
}
int main() {
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  while (m--) {
    int s, x, y, z;
    scanf("%d", &s);
    if (s == 1) {
      scanf("%d%d%d", &x, &y, &z);
      update(1, x, y, z);
    }
    if (s == 2) {
      scanf("%d%d", &x, &y);
      printf("%lld\n", ask(1, x, y));
    }
  }
}
