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 mark[100007 * 4], delta[100007 * 4], sum[100007 * 4];
int n, m, v, l, r;
void build(int x, int tl, int tr) {
  if (tl < tr) {
    int mid = (tl + tr) >> 1;
    build(x * 2, tl, mid);
    build(x * 2 + 1, mid + 1, tr);
  } else
    mark[x] = tl;
}
void work(int x, int tl, int tr) {
  int ch1 = x * 2, ch2 = ch1 + 1;
  if (mark[x] > 0) {
    delta[x] += abs(mark[x] - v);
    sum[x] += abs(mark[x] - v) * (tr - tl + 1);
    mark[x] = 0;
  } else {
    int ch1 = x * 2, ch2 = ch1 + 1;
    int mid = (tl + tr) >> 1;
    work(ch1, tl, mid);
    work(ch2, mid + 1, tr);
    sum[x] = sum[ch1] + sum[ch2] + delta[x] * (tr - tl + 1);
  }
}
void modify(int x, int tl, int tr) {
  if (l <= tl && tr <= r) {
    work(x, tl, tr);
    mark[x] = v;
  } else {
    int ch1 = x * 2, ch2 = ch1 + 1;
    if (mark[x] > 0) mark[ch1] = mark[ch2] = mark[x], mark[x] = 0;
    int mid = (tl + tr) >> 1;
    if (l <= mid) modify(ch1, tl, mid);
    if (mid < r) modify(ch2, mid + 1, tr);
    sum[x] = sum[ch1] + sum[ch2] + delta[x] * (tr - tl + 1);
  }
}
long long query(int x, int tl, int tr) {
  if (l <= tl && tr <= r)
    return sum[x];
  else {
    long long ans = (min(r, tr) - max(l, tl) + 1) * delta[x];
    int mid = (tl + tr) >> 1;
    if (l <= mid) ans += query(x * 2, tl, mid);
    if (mid < r) ans += query(x * 2 + 1, mid + 1, tr);
    return ans;
  }
}
int main() {
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  int type;
  while (m--) {
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d%d", &l, &r, &v);
      modify(1, 1, n);
    } else {
      scanf("%d%d", &l, &r);
      printf("%I64d\n", query(1, 1, n));
    }
  }
  return 0;
}
