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 = 1 << 17;
long long n, m;
long long la[N * 2 + 10], c[N * 2 + 10], s[N * 2 + 10];
void lp(long long i, long long v) {
  la[i << 1] += la[i];
  s[i << 1] += la[i] * v;
  la[i << 1 | 1] += la[i];
  s[i << 1 | 1] += la[i] * v;
  if (c[i]) c[i << 1] = c[i << 1 | 1] = c[i];
  la[i] = 0;
}
long long q(long long l, long long r, long long L, long long R, long long i) {
  if (r < L || l > R) return 0;
  if (l >= L && r <= R) return s[i];
  lp(i, (r - l + 1) >> 1);
  long long m = (l + r) >> 1, rt = 0;
  rt += q(l, m, L, R, i << 1);
  rt += q(m + 1, r, L, R, i << 1 | 1);
  return rt;
}
void u(long long l, long long r, long long L, long long R, long long x,
       long long i) {
  if (r < L || l > R) return;
  if (l >= L && r <= R && c[i]) {
    s[i] += abs(c[i] - x) * (r - l + 1);
    la[i] += abs(c[i] - x);
    c[i] = x;
    return;
  }
  lp(i, (r - l + 1) >> 1);
  long long m = (l + r) >> 1;
  u(l, m, L, R, x, i << 1);
  u(m + 1, r, L, R, x, i << 1 | 1);
  if (c[i << 1] == c[i << 1 | 1])
    c[i] = c[i << 1];
  else
    c[i] = 0;
  s[i] = s[i << 1] + s[i << 1 | 1];
}
int main() {
  scanf("%I64d %I64d", &n, &m);
  for (long long i = 1; i <= n; i++) c[N + i - 1] = i;
  while (m--) {
    long long t, l, r, x;
    scanf("%I64d %I64d %I64d", &t, &l, &r);
    if (t == 1) {
      scanf("%I64d", &x);
      u(1, N, l, r, x, 1);
    } else
      printf("%I64d\n", q(1, N, l, r, 1));
  }
  return 0;
}
