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 n, m, t, l, r, x, mark[410000];
long long delta[410000], sum[410000];
void read(int &x) {
  char k;
  for (k = getchar(); k <= 32; k = getchar())
    ;
  for (x = 0; '0' <= k; k = getchar()) x = x * 10 + k - '0';
}
void mkt(int k, int q, int h) {
  if (q < h) {
    mkt(k * 2, q, (q + h) / 2);
    mkt(k * 2 + 1, (q + h) / 2 + 1, h);
    mark[k] = 0;
  } else
    mark[k] = q;
}
long long query(int k, int q, int h, int l, int r) {
  if (l <= q && h <= r) return sum[k];
  if (r <= (q + h) / 2)
    return query(k * 2, q, (q + h) / 2, l, r) + 1LL * delta[k] * (r - l + 1);
  if ((q + h) / 2 < l)
    return query(k * 2 + 1, (q + h) / 2 + 1, h, l, r) +
           1LL * delta[k] * (r - l + 1);
  return query(k * 2, q, (q + h) / 2, l, (q + h) / 2) +
         query(k * 2 + 1, (q + h) / 2 + 1, h, (q + h) / 2 + 1, r) +
         1LL * delta[k] * (r - l + 1);
}
void Clear(int k, int q, int h) {
  if (mark[k] > 0) {
    delta[k] += abs(mark[k] - x);
    sum[k] += 1LL * abs(mark[k] - x) * (h - q + 1);
  } else {
    Clear(k * 2, q, (q + h) / 2);
    Clear(k * 2 + 1, (q + h) / 2 + 1, h);
    sum[k] = sum[k * 2] + sum[k * 2 + 1] + 1LL * delta[k] * (h - q + 1);
  }
  mark[k] = -1;
}
void modify(int k, int q, int h) {
  if (l <= q && h <= r)
    Clear(k, q, h), mark[k] = x;
  else {
    if (mark[k] > 0) mark[k * 2] = mark[k * 2 + 1] = mark[k], mark[k] = 0;
    if (r <= (q + h) / 2)
      modify(k * 2, q, (q + h) / 2);
    else if ((q + h) / 2 < l)
      modify(k * 2 + 1, (q + h) / 2 + 1, h);
    else
      modify(k * 2, q, (q + h) / 2), modify(k * 2 + 1, (q + h) / 2 + 1, h);
    mark[k] = 0;
    sum[k] = sum[k * 2] + sum[k * 2 + 1] + 1LL * delta[k] * (h - q + 1);
  }
}
int main() {
  read(n);
  read(m);
  mkt(1, 1, n);
  while (m--) {
    read(t);
    read(l);
    read(r);
    if (t == 1) {
      read(x);
      modify(1, 1, n);
    } else {
      printf("%I64d\n", query(1, 1, n, l, r));
    }
  }
}
