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 = 112345;
long long t[4 * N];
long long lzs[4 * N], lzm[4 * N];
int a[4 * N], b[4 * N];
void build(int i, int l, int r) {
  if (l == r) {
    a[i] = b[i] = l;
  } else {
    int m = (l + r) / 2;
    build(2 * i, l, m);
    build(2 * i + 1, m + 1, r);
    a[i] = l;
    b[i] = r;
  }
}
void push(int i, int l, int r) {
  if (lzs[i]) {
    t[i] += 1ll * lzs[i] * (r - l + 1);
    if (l != r) {
      lzs[2 * i] += lzs[i];
      lzs[2 * i + 1] += lzs[i];
    }
    lzs[i] = 0;
  }
  if (lzm[i]) {
    a[i] = b[i] = lzm[i];
    if (l != r) {
      lzm[2 * i] = lzm[i];
      lzm[2 * i + 1] = lzm[i];
    }
    lzm[i] = 0;
  }
}
long long query(int i, int l, int r, int ql, int qr) {
  push(i, l, r);
  if (ql <= l && r <= qr) return t[i];
  if (qr < l || r < ql) return 0;
  int m = (l + r) / 2;
  return query(2 * i, l, m, ql, qr) + query(2 * i + 1, m + 1, r, ql, qr);
}
void update(int i, int l, int r, int ql, int qr, int x) {
  push(i, l, r);
  if (qr < l || r < ql) return;
  if (ql <= l && r <= qr && a[i] == b[i]) {
    lzs[i] += abs(x - a[i]);
    lzm[i] = x;
    push(i, l, r);
  } else {
    int m = (l + r) / 2;
    update(2 * i, l, m, ql, qr, x);
    update(2 * i + 1, m + 1, r, ql, qr, x);
    t[i] = t[2 * i] + t[2 * i + 1];
    a[i] = min(a[2 * i], a[2 * i + 1]);
    b[i] = max(b[2 * i], b[2 * i + 1]);
  }
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  while (m--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l, r, x;
      scanf("%d%d%d", &l, &r, &x);
      update(1, 1, n, l, r, x);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%lld\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}
