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 = 100002;
int color[Maxn << 2];
long long sum[Maxn << 2], add[Maxn << 2];
void build(int l, int r, int x) {
  if (l == r) {
    color[x] = l;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, x << 1), build(mid + 1, r, x << 1 | 1);
}
void down(int l, int r, int x) {
  int mid = (l + r) >> 1;
  color[x << 1] = color[x << 1 | 1] = color[x];
  if (!add[x]) return;
  add[x << 1] += add[x];
  add[x << 1 | 1] += add[x];
  sum[x << 1] += add[x] * (mid - l + 1);
  sum[x << 1 | 1] += add[x] * (r - mid);
  add[x] = 0;
}
void merge(int l, int r, int x) {
  color[x] = color[x << 1] == color[x << 1 | 1] ? color[x << 1] : 0;
  sum[x] = sum[x << 1] + sum[x << 1 | 1];
}
void up(int L, int R, int w, int l, int r, int x) {
  int mid = (l + r) >> 1;
  if (L <= l && R >= r) {
    if (color[x]) {
      sum[x] += (r - l + ((long long)1)) * abs(w - color[x]);
      add[x] += abs(w - color[x]);
      color[x] = w;
    } else {
      up(L, R, w, l, mid, x << 1), up(L, R, w, mid + 1, r, x << 1 | 1);
      merge(l, r, x);
    }
    return;
  }
  if (color[x]) down(l, r, x);
  if (L <= mid) up(L, R, w, l, mid, x << 1);
  if (R > mid) up(L, R, w, mid + 1, r, x << 1 | 1);
  merge(l, r, x);
}
long long query(int L, int R, int l, int r, int x) {
  if (L <= l && R >= r) return sum[x];
  if (color[x]) down(l, r, x);
  long long ret = 0;
  int mid = (l + r) >> 1;
  if (L <= mid) ret += query(L, R, l, mid, x << 1);
  if (R > mid) ret += query(L, R, mid + 1, r, x << 1 | 1);
  return ret;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  build(1, n, 1);
  while (m--) {
    int a, b, c, d;
    scanf("%d%d%d", &a, &b, &c);
    if (a == 1) {
      scanf("%d", &d);
      up(b, c, d, 1, n, 1);
    } else
      printf("%I64d\n", query(b, c, 1, n, 1));
  }
}
