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 int maxn = 100005;
long long add[400], addb[maxn], a[maxn], sum[400], b[maxn];
int L[400], R[400], belong[maxn];
int n, m, block, cnt;
void init() {
  block = sqrt(n);
  if (n % block)
    cnt = n / block + 1;
  else
    cnt = n / block;
  for (int i = 1; i <= cnt; i++) {
    L[i] = (i - 1) * block + 1;
    R[i] = i * block;
  }
  R[cnt] = n;
  for (int i = 1; i <= n; i++) {
    belong[i] = (i - 1) / block + 1;
  }
}
void push_down(int x) {
  if (add[x] != 0) {
    for (int i = L[x]; i <= R[x]; i++) {
      b[i] += addb[x];
      a[i] = add[x];
    }
    add[x] = addb[x] = 0;
  }
}
void update(int l, int r, int x) {
  int ll = belong[l];
  int rr = belong[r];
  if (ll == rr) {
    push_down(ll);
    for (int i = l; i <= r; i++) {
      sum[ll] += abs(x - a[i]);
      b[i] += abs(x - a[i]);
      a[i] = x;
    }
  } else {
    push_down(ll);
    push_down(rr);
    for (int i = l; i <= R[ll]; i++) {
      sum[ll] += abs(x - a[i]);
      b[i] += abs(x - a[i]);
      a[i] = x;
    }
    for (int i = L[rr]; i <= r; i++) {
      sum[rr] += abs(x - a[i]);
      b[i] += abs(x - a[i]);
      a[i] = x;
    }
    for (int i = ll + 1; i <= rr - 1; i++) {
      if (add[i] != 0) {
        sum[i] += abs(x - add[i]) * (R[i] - L[i] + 1);
        addb[i] += abs(x - add[i]);
        add[i] = x;
      } else {
        add[i] = x;
        for (int j = L[i]; j <= R[i]; j++) {
          b[j] += abs(x - a[j]);
          sum[i] += abs(x - a[j]);
        }
      }
    }
  }
}
long long ask(int l, int r) {
  int ll = belong[l];
  int rr = belong[r];
  long long ans = 0;
  if (ll == rr) {
    for (int i = l; i <= r; i++) {
      ans += b[i] + addb[ll];
    }
  } else {
    for (int i = ll + 1; i <= rr - 1; i++) {
      ans += sum[i];
    }
    for (int i = l; i <= R[ll]; i++) {
      ans += b[i] + addb[ll];
    }
    for (int i = L[rr]; i <= r; i++) {
      ans += b[i] + addb[rr];
    }
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    a[i] = i;
  }
  init();
  while (m--) {
    int f, l, r, x;
    scanf("%d", &f);
    if (f == 1) {
      scanf("%d%d%d", &l, &r, &x);
      update(l, r, x);
    } else {
      scanf("%d%d", &l, &r);
      printf("%lld\n", ask(l, r));
    }
  }
  return 0;
}
