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;
int t, l, r, x;
int color[4 * 100002];
long long sum[4 * 100002], delta[4 * 100002];
void init(int root, int l, int r) {
  if (l + 1 == r) {
    color[root] = l;
    return;
  }
  init(root * 2, l, (l + r) / 2);
  init(root * 2 + 1, (l + r) / 2, r);
}
void clear(int root, int a, int b, int x) {
  if (color[root]) {
    long long diff = abs(color[root] - x);
    sum[root] += diff * (b - a);
    delta[root] += diff;
    color[root] = 0;
  } else {
    clear(root * 2, a, (a + b) / 2, x);
    clear(root * 2 + 1, (a + b) / 2, b, x);
    sum[root] = sum[root * 2] + sum[root * 2 + 1] + delta[root] * (b - a);
  }
}
void paint(int root, int a, int b, int l, int r, int x) {
  if (l <= a && r >= b) {
    clear(root, a, b, x);
    color[root] = x;
    return;
  }
  if (color[root]) {
    color[root * 2] = color[root * 2 + 1] = color[root];
  }
  color[root] = 0;
  if (l >= (a + b) / 2)
    paint(root * 2 + 1, (a + b) / 2, b, l, r, x);
  else if (r <= (a + b) / 2)
    paint(root * 2, a, (a + b) / 2, l, r, x);
  else
    paint(root * 2, a, (a + b) / 2, l, r, x),
        paint(root * 2 + 1, (a + b) / 2, b, l, r, x);
  sum[root] = sum[root * 2] + sum[root * 2 + 1] + delta[root] * (b - a);
}
long long getSum(int root, int a, int b, int l, int r) {
  if (l <= a && r >= b) return sum[root];
  long long d = delta[root] * (min(b, r) - max(l, a));
  if (l >= (a + b) / 2)
    return d + getSum(root * 2 + 1, (a + b) / 2, b, l, r);
  else if (r <= (a + b) / 2)
    return getSum(root * 2, a, (a + b) / 2, l, r) + d;
  else
    return d + getSum(root * 2, a, (a + b) / 2, l, r) +
           getSum(root * 2 + 1, (a + b) / 2, b, l, r);
}
int main() {
  scanf("%d%d", &n, &m);
  init(1, 1, n + 1);
  while (m--) {
    scanf("%d%d%d", &t, &l, &r);
    r++;
    if (t == 1) {
      scanf("%d", &x);
      paint(1, 1, n + 1, l, r, x);
    } else {
      printf("%I64d\n", getSum(1, 1, n + 1, l, r));
    }
  }
  return 0;
}
