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;
long long sum[102400 << 2], sub[102400 << 2], mark[102400 << 2];
int lp[102400 << 2], rp[102400 << 2];
void build(int u, int l, int r) {
  lp[u] = l;
  rp[u] = r;
  if (l < r) {
    build(u << 1, l, (l + r) >> 1);
    build(u << 1 | 1, ((l + r) >> 1) + 1, r);
  } else
    mark[u] = l;
}
void update(int u) {
  sum[u] =
      sum[u << 1] + sum[u << 1 | 1] + (long long)(rp[u] - lp[u] + 1) * sub[u];
}
long long query(int u, int l, int r) {
  if (rp[u] < l || lp[u] > r) return 0;
  if (l <= lp[u] && rp[u] <= r) return sum[u];
  int m = (lp[u] + rp[u]) >> 1;
  return query(u << 1, l, r) + query(u << 1 | 1, l, r) +
         max(min(r, rp[u]) - max(l, lp[u]) + 1, 0) * sub[u];
}
void clear(int u, int x) {
  if (mark[u]) {
    sub[u] += abs(mark[u] - (long long)x);
    sum[u] += (long long)(rp[u] - lp[u] + 1) * abs(mark[u] - (long long)x);
    mark[u] = 0;
  } else {
    if (lp[u] == rp[u]) return;
    clear(u << 1, x);
    clear(u << 1 | 1, x);
    update(u);
  }
}
void change(int u, int l, int r, int x) {
  if (r < lp[u] || l > rp[u]) return;
  if (l <= lp[u] && rp[u] <= r) {
    clear(u, x);
    mark[u] = x;
    return;
  }
  if (mark[u]) {
    mark[u << 1] = mark[u << 1 | 1] = mark[u];
    mark[u] = 0;
  }
  change(u << 1, l, r, x);
  change(u << 1 | 1, l, r, x);
  update(u);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  for (int i = 0; i < m; i++) {
    int tp, l, r, c;
    scanf("%d%d%d", &tp, &l, &r);
    if (tp - 1)
      cout << query(1, l, r) << endl;
    else {
      scanf("%d", &c);
      change(1, l, r, c);
    }
  }
  return 0;
}
