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>
#pragma GCC optimize("Ofast", "no-stack-protector")
using namespace std;
int n, m;
int type, l, r, x;
int a[100000], bk[100000], chg[400], bs;
long long int v[100000], sum[400], add[400];
void push(int k) {
  if (!chg[k]) return;
  for (int i = k * bs; i < (k + 1) * bs; ++i) {
    v[i] += add[k], a[i] = chg[k];
  }
  add[k] = chg[k] = 0;
}
void brute(int l, int r, int x) {
  for (int i = l; i <= r; ++i) {
    v[i] += abs(x - a[i]), sum[bk[i]] += abs(x - a[i]);
    a[i] = x;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m, bs = ceil(sqrt(n));
  for (int i = 0; i < n; ++i) a[i] = i + 1, bk[i] = i / bs;
  while (m--) {
    cin >> type >> l >> r, --l, --r;
    if (type == 1) {
      cin >> x;
      push(bk[l]), push(bk[r]);
      if (bk[l] == bk[r]) {
        brute(l, r, x);
      } else {
        brute(l, (bk[l] + 1) * bs - 1, x);
        brute(bk[r] * bs, r, x);
        l = bk[l] + 1, r = bk[r] - 1;
        while (l <= r) {
          if (!chg[l])
            brute(l * bs, (l + 1) * bs - 1, x);
          else
            add[l] += abs(x - chg[l]),
                sum[l] += (long long int)abs(x - chg[l]) * bs;
          chg[l++] = x;
        }
      }
    } else {
      long long int ans = 0;
      push(bk[l]), push(bk[r]);
      if (bk[l] == bk[r]) {
        for (int i = l; i <= r; ++i) ans += v[i];
      } else {
        for (int i = l; i < (bk[l] + 1) * bs; ++i) ans += v[i];
        for (int i = bk[r] * bs; i <= r; ++i) ans += v[i];
        l = bk[l] + 1, r = bk[r] - 1;
        while (l <= r) ans += sum[l], ++l;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
