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 = (int)1e6 + 5;
long long add[N << 2], sum[N << 2], color[N << 1];
int n, m;
void down(int rt, int len) {
  if (add[rt]) {
    add[(rt << 1)] += add[rt];
    add[(rt << 1 | 1)] += add[rt];
    sum[(rt << 1)] += add[rt] * (len - (len >> 1));
    sum[(rt << 1 | 1)] += add[rt] * (len >> 1);
    add[rt] = 0;
    color[(rt << 1)] = color[(rt << 1 | 1)] = color[rt];
    color[rt] = 0;
  }
}
void build(int rt, int l, int r) {
  if (l == r) {
    color[rt] = l;
    return;
  }
  int mid = (l + r) >> 1;
  build((rt << 1), l, mid);
  build((rt << 1 | 1), mid + 1, r);
  color[rt] = (color[(rt << 1)] == color[(rt << 1 | 1)]) ? color[(rt << 1)] : 0;
  sum[rt] = sum[(rt << 1)] + sum[(rt << 1 | 1)];
}
void modify(int rt, int l, int r, int L, int R, int x) {
  if (L <= l && r <= R && color[rt]) {
    add[rt] += abs(color[rt] - x);
    sum[rt] += (r - l + 1) * abs(color[rt] - x);
    color[rt] = x;
    return;
  }
  down(rt, r - l + 1);
  int mid = (l + r) >> 1;
  if (L <= mid) modify((rt << 1), l, mid, L, R, x);
  if (mid < R) modify((rt << 1 | 1), mid + 1, r, L, R, x);
  color[rt] = (color[(rt << 1)] == color[(rt << 1 | 1)]) ? color[(rt << 1)] : 0;
  sum[rt] = sum[(rt << 1)] + sum[(rt << 1 | 1)];
}
long long query(int rt, int l, int r, int L, int R) {
  if (L <= l && r <= R) {
    return sum[rt];
  }
  down(rt, r - l + 1);
  long long t = 0ll;
  int mid = (l + r) >> 1;
  if (L <= mid) t += query((rt << 1), l, mid, L, R);
  if (mid < R) t += query((rt << 1 | 1), mid + 1, r, L, R);
  return t;
}
int main() {
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  while (m--) {
    int op, l, r, x;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d%d", &l, &r, &x);
      modify(1, 1, n, l, r, x);
    } else {
      scanf("%d%d", &l, &r);
      printf("%I64d\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}
