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 = 1e5 + 10;
long long a[maxn], b[maxn];
int belong[maxn];
int L[maxn], R[maxn];
int block;
int num;
long long sum[maxn];
int n, q;
long long flag[maxn], flag2[maxn];
void init() {
  for (int i = 1; i <= n; ++i) a[i] = i;
  int num = sqrt(n);
  block = n / num;
  if (n % block) ++num;
  for (int i = 1; i <= num; ++i) {
    L[i] = (i - 1) * block + 1;
    R[i] = i * block;
  }
  R[n] = num;
  for (int i = 1; i <= n; ++i) {
    belong[i] = (i - 1) / block + 1;
  }
}
long long aaa(long long x) {
  if (x < 0) return -x;
  return x;
}
void update(int l, int r, int x) {
  for (int i = l; i <= r;) {
    int id = belong[i];
    if (i == L[id] && r >= R[id]) {
      if (flag[id]) {
        sum[id] += (long long)(R[id] - L[id] + 1) * aaa(x - flag[id]);
        flag2[id] += aaa(x - flag[id]);
        flag[id] = x;
      } else {
        flag[id] = x;
        for (int j = L[id]; j <= R[id]; ++j) {
          sum[id] += aaa(a[j] - x);
          b[j] += aaa(a[j] - x);
          a[j] = x;
        }
      }
      i = R[id] + 1;
    } else {
      if (flag[id]) {
        for (int j = L[id]; j <= R[id]; ++j) a[j] = flag[id];
        flag[id] = 0;
      }
      sum[id] += aaa(a[i] - x);
      b[i] += aaa(a[i] - x);
      a[i] = x;
      ++i;
    }
  }
}
long long query(int l, int r) {
  long long ans = 0LL;
  for (int i = l; i <= r;) {
    int id = belong[i];
    if (i == L[id] && r >= R[id]) {
      ans += sum[id];
      i = R[id] + 1;
    } else {
      ans += b[i] + flag2[id];
      ++i;
    }
  }
  return ans;
}
int main() {
  scanf("%d %d", &n, &q);
  init();
  int op;
  while (q--) {
    scanf("%d", &op);
    if (op == 1) {
      int l, r, v;
      scanf("%d %d %d", &l, &r, &v);
      update(l, r, v);
    } else {
      int l, r;
      scanf("%d %d", &l, &r);
      printf("%lld\n", query(l, r));
    }
  }
  return 0;
}
