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 long long N = 1e5 + 7, magic = 320;
long long a[N], ans[N], prop[magic], st[magic], en[magic], lzy[magic],
    block_ans[N];
void apply(long long k) {
  if (lzy[k]) {
    for (long long i = st[k]; i <= en[k]; i++) {
      ans[i] += prop[k];
      a[i] = lzy[k];
    }
    prop[k] = lzy[k] = 0;
  }
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    long long k = i / magic;
    en[k] = i;
    st[k] = !st[k] ? i : st[k];
    a[i] = i;
  }
  for (long long i = 1; i <= m; i++) {
    long long typ;
    cin >> typ;
    if (typ == 1) {
      long long l, r, x;
      cin >> l >> r >> x;
      for (long long j = l; j <= r; j++) {
        long long k = j / magic;
        if (j != st[k] || en[k] > r) {
          apply(k);
          for (long long lx = j; lx <= min(en[k], r); lx++) {
            block_ans[k] += abs(a[lx] - x);
            ans[lx] += abs(a[lx] - x);
            a[lx] = x;
          }
        } else {
          if (!lzy[k]) {
            lzy[k] = x;
            for (long long lx = st[k]; lx <= en[k]; ++lx) {
              block_ans[k] += abs(a[lx] - x);
              ans[lx] += abs(a[lx] - x);
            }
          } else {
            prop[k] += abs(lzy[k] - x);
            block_ans[k] += abs(lzy[k] - x) * (en[k] - st[k] + 1);
            lzy[k] = x;
          }
        }
        j = en[k];
      }
    } else {
      long long l, r, s = 0;
      cin >> l >> r;
      for (long long j = l; j <= r; j++) {
        long long k = j / magic;
        if (j != st[k] || en[k] > r) {
          apply(k);
          for (long long lx = j; lx <= min(en[k], r); lx++) s += ans[lx];
        } else {
          s += block_ans[k];
        }
        j = en[k];
      }
      cout << s << "\n";
    }
  }
}
