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;
struct tree {
  long long sum, add, col;
};
tree seg[411111];
void build(long long tren, long long l, long long d) {
  if (l > d) return;
  if (d == l) {
    seg[tren].sum = 0;
    seg[tren].col = l;
    seg[tren].add = 0;
    return;
  }
  build(tren * 2, l, (l + d) / 2);
  build(tren * 2 + 1, 1 + (l + d) / 2, d);
  seg[tren].sum = seg[tren * 2].sum + seg[tren * 2 + 1].sum;
  seg[tren].add = 0;
  if (seg[tren * 2].col == seg[tren * 2 + 1].col)
    seg[tren].col = seg[tren * 2].col;
  else
    seg[tren].col = -1;
}
void update(long long tren, long long l, long long d, long long i, long long j,
            long long x) {
  if (l > d || d < i || l > j) return;
  if (i <= l && d <= j) {
    if (seg[tren].col > 0) {
      seg[tren].sum += abs(seg[tren].col - x) * (d - l + 1);
      seg[tren].add += abs(seg[tren].col - x);
      seg[tren].col = x;
      return;
    }
    update(tren * 2, l, (l + d) / 2, i, j, x);
    update(tren * 2 + 1, 1 + (l + d) / 2, d, i, j, x);
    seg[tren].sum = seg[tren * 2].sum + seg[tren * 2 + 1].sum;
    seg[tren].col = x;
    return;
  }
  if (l != d && seg[tren].col > 0) {
    long long mid = (l + d) / 2;
    seg[tren * 2].sum += seg[tren].add * (mid - l + 1);
    seg[tren * 2].add += seg[tren].add;
    seg[tren * 2].col = seg[tren].col;
    seg[tren * 2 + 1].sum += seg[tren].add * (d - mid);
    seg[tren * 2 + 1].add += seg[tren].add;
    seg[tren * 2 + 1].col = seg[tren].col;
    seg[tren].add = 0;
  }
  update(tren * 2, l, (l + d) / 2, i, j, x);
  update(tren * 2 + 1, 1 + (l + d) / 2, d, i, j, x);
  seg[tren].sum = seg[tren * 2].sum + seg[tren * 2 + 1].sum;
  if (seg[tren * 2].col == seg[tren * 2 + 1].col)
    seg[tren].col = seg[tren * 2].col;
  else
    seg[tren].col = -1;
}
long long query(long long tren, long long l, long long d, long long i,
                long long j) {
  if (l > d || d < i || l > j) return 0;
  if (i <= l && d <= j) return seg[tren].sum;
  if (l != d && seg[tren].col > 0) {
    long long mid = (l + d) / 2;
    seg[tren * 2].sum += seg[tren].add * (mid - l + 1);
    seg[tren * 2].add += seg[tren].add;
    seg[tren * 2].col = seg[tren].col;
    seg[tren * 2 + 1].sum += seg[tren].add * (d - mid);
    seg[tren * 2 + 1].add += seg[tren].add;
    seg[tren * 2 + 1].col = seg[tren].col;
    seg[tren].add = 0;
  }
  long long left = query(tren * 2, l, (l + d) / 2, i, j);
  long long right = query(tren * 2 + 1, 1 + (l + d) / 2, d, i, j);
  return (left + right);
}
int main() {
  long long n, m;
  scanf("%lld%lld", &n, &m);
  build(1, 1, n);
  while (m--) {
    long long t, l, r, x;
    scanf("%lld", &t);
    if (t == 1) {
      scanf("%lld%lld%lld", &l, &r, &x);
      update(1, 1, n, l, r, x);
    } else {
      scanf("%lld%lld", &l, &r);
      printf("%lld\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}
