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;
int col[101100 << 2];
long long val[101100 << 2];
long long sum[101100 << 2];
void push_up(int root) {
  sum[root] = sum[root << 1] + sum[root << 1 | 1];
  if (col[root << 1] == col[root << 1 | 1])
    col[root] = col[root << 1];
  else
    col[root] = 0;
}
void push_down(int root, int l, int r) {
  if (val[root]) {
    int m = (l + r) >> 1;
    sum[root << 1] += val[root] * (m - l + 1);
    sum[root << 1 | 1] += val[root] * (r - m);
    val[root << 1] += val[root], val[root << 1 | 1] += val[root];
    col[root << 1] = col[root << 1 | 1] = col[root];
    val[root] = 0;
    col[root] = 0;
  }
}
void build(int l, int r, int root) {
  col[root] = sum[root] = val[root] = 0;
  if (l == r) {
    col[root] = l;
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, root << 1);
  build(m + 1, r, root << 1 | 1);
  push_up(root);
}
void update(int l, int r, int root, int L, int R, int a) {
  if (L <= l && r <= R && col[root]) {
    sum[root] += fabs(col[root] - a) * (r - l + 1);
    val[root] += fabs(col[root] - a);
    col[root] = a;
    return;
  }
  int m = (l + r) >> 1;
  push_down(root, l, r);
  if (L <= m) update(l, m, root << 1, L, R, a);
  if (R > m) update(m + 1, r, root << 1 | 1, L, R, a);
  push_up(root);
}
long long query(int l, int r, int root, int L, int R) {
  if (L <= l && r <= R) return sum[root];
  int m = (l + r) >> 1;
  push_down(root, l, r);
  long long res = 0;
  if (L <= m) res += query(l, m, root << 1, L, R);
  if (R > m) res += query(m + 1, r, root << 1 | 1, L, R);
  push_up(root);
  return res;
}
int main() {
  int a, b, c, d;
  int i, j, k;
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    build(1, n, 1);
    for (i = 0; i < (m); ++i) {
      scanf("%d", &a);
      if (a == 1) {
        scanf("%d%d%d", &b, &c, &d);
        update(1, n, 1, b, c, d);
      } else {
        scanf("%d%d", &b, &c);
        printf("%I64d\n", query(1, n, 1, b, c));
      }
    }
  }
  return 0;
}
