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>
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
const int root = 1e8;
using namespace std;
long long n, m, k;
long long block, num, belong[maxn], l[1000], r[1000], a[maxn], lazy[400],
    lazb[400], b[maxn];
long long sum[400];
void built() {
  for (int i = 1; i <= n; i++) a[i] = i;
  block = sqrt(n);
  num = n / block;
  if (n % block) num++;
  for (int i = 1; i <= n; i++) belong[i] = (i - 1) / block + 1;
  for (int i = 1; i <= num; i++) l[i] = (i - 1) * block + 1, r[i] = i * block;
  r[num] = n;
}
void update(long long x, long long y, long long k) {
  int f1 = belong[x], f2 = belong[y];
  if (f1 == f2) {
    if (lazy[f1]) {
      for (int i = l[f1]; i <= r[f1]; i++) {
        a[i] = lazy[f1];
      }
      lazy[f1] = 0;
      for (int i = x; i <= y; i++) {
        b[i] += abs(k - a[i]);
        sum[f1] += abs(k - a[i]);
        a[i] = k;
      }
      return;
    } else {
      for (int i = x; i <= y; i++) {
        b[i] += abs(a[i] - k);
        sum[f1] += abs(a[i] - k);
        a[i] = k;
      }
      return;
    }
  }
  if (lazy[f1]) {
    for (int i = l[f1]; i <= r[f1]; i++) {
      a[i] = lazy[f1];
    }
    lazy[f1] = 0;
  }
  for (int i = x; i <= r[f1]; i++) {
    b[i] += abs(k - a[i]);
    sum[f1] += abs(k - a[i]);
    a[i] = k;
  }
  for (int i = f1 + 1; i < f2; i++) {
    if (lazy[i]) {
      lazb[i] += abs(k - lazy[i]);
      sum[i] += abs(k - lazy[i]) * (r[i] - l[i] + 1);
      lazy[i] = k;
    } else {
      for (int j = l[i]; j <= r[i]; j++) {
        sum[i] += abs(a[j] - k);
        b[j] += abs(a[j] - k);
      }
      lazy[i] = k;
    }
  }
  if (lazy[f2]) {
    for (int i = l[f2]; i <= r[f2]; i++) a[i] = lazy[f2];
    lazy[f2] = 0;
  }
  for (int i = l[f2]; i <= y; i++) {
    b[i] += abs(a[i] - k);
    sum[f2] += abs(a[i] - k);
    a[i] = k;
  }
  return;
}
void slove(int x, int y) {
  long long ans = 0;
  int f1 = belong[x], f2 = belong[y];
  if (f1 == f2) {
    for (int i = x; i <= y; i++) {
      ans += (b[i] + lazb[f1]);
    }
    printf("%lld\n", ans);
    return;
  }
  for (int i = x; i <= r[f1]; i++) {
    ans += (b[i] + lazb[f1]);
  }
  for (int i = f1 + 1; i < f2; i++) {
    ans += sum[i];
  }
  for (int i = l[f2]; i <= y; i++) {
    ans += (b[i] + lazb[f2]);
  }
  printf("%lld\n", ans);
  return;
}
int main() {
  scanf("%d %d", &n, &m);
  built();
  while (m--) {
    int op, x, y, k;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d %d %d", &x, &y, &k);
      update(x, y, k);
    } else {
      scanf("%d %d", &x, &y);
      slove(x, y);
    }
  }
  return 0;
}
