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 maxn = 1e5 + 5;
long long color[maxn << 2], seg[maxn << 2], val[maxn << 2];
void build(int l, int r, int pos) {
  seg[pos] = val[pos] = 0;
  color[pos] = -1;
  if (l == r) {
    color[pos] = l;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, pos << 1);
  build(mid + 1, r, pos << 1 | 1);
}
void push_down(int l, int r, int pos) {
  if (color[pos] != -1) {
    int mid = (l + r) >> 1;
    seg[pos << 1] += val[pos] * (mid - l + 1);
    seg[pos << 1 | 1] += val[pos] * (r - mid);
    val[pos << 1] += val[pos];
    val[pos << 1 | 1] += val[pos];
    color[pos << 1] = color[pos];
    color[pos << 1 | 1] = color[pos];
    color[pos] = -1;
    val[pos] = 0;
  }
}
void push_up(int l, int r, int pos) {
  if (color[pos << 1] == color[pos << 1 | 1])
    color[pos] = color[pos << 1];
  else
    color[pos] = -1;
  seg[pos] = seg[pos << 1] + seg[pos << 1 | 1];
}
void update(int l, int r, int pos, int ua, int ub, long long x) {
  if (ua <= l and ub >= r) {
    if (color[pos] != -1) {
      seg[pos] += abs(1LL * x - color[pos]) * (r - l + 1);
      val[pos] += abs(1LL * x - color[pos]);
      color[pos] = x;
      return;
    }
  }
  int mid = (l + r) >> 1;
  push_down(l, r, pos);
  if (ua <= mid) update(l, mid, pos << 1, ua, ub, x);
  if (ub > mid) update(mid + 1, r, pos << 1 | 1, ua, ub, x);
  push_up(l, r, pos);
}
long long query(int l, int r, int pos, int ua, int ub) {
  if (ua <= l and ub >= r) return seg[pos];
  push_down(l, r, pos);
  int mid = (l + r) >> 1;
  long long res = 0;
  if (ua <= mid) res += query(l, mid, pos << 1, ua, ub);
  if (ub > mid) res += query(mid + 1, r, pos << 1 | 1, ua, ub);
  return res;
}
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    build(1, n, 1);
    while (m--) {
      int type, l, r;
      long long x;
      scanf("%d%d%d", &type, &l, &r);
      if (type == 1) {
        scanf("%I64d", &x);
        update(1, n, 1, l, r, x);
      } else {
        printf("%I64d\n", query(1, n, 1, l, r));
      }
    }
  }
  return 0;
}
