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;
template <typename T>
inline T read(T& x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x = f ? -x : x;
}
const int N = 1e5 + 10;
int n, m, sz, i, op, l, r, x, flag[N], block[N], color[N];
long long val[N], sum[N], tag[N];
void reset(int x) {
  if (flag[x] == -1) return;
  for (int i = (x - 1) * sz + 1; i <= (x * sz > n ? n : x * sz); i++) {
    color[i] = flag[x];
  }
  flag[x] = -1;
}
void update(int a, int b, int x) {
  int i, j;
  reset(block[a]);
  for (i = a; i <= (block[a] * sz > b ? b : block[a] * sz); i++) {
    val[i] += abs(color[i] - x);
    sum[block[i]] += abs(color[i] - x);
    color[i] = x;
  }
  if (block[a] != block[b]) {
    reset(block[b]);
    for (i = (block[b] - 1) * sz + 1; i <= b; i++) {
      val[i] += abs(color[i] - x);
      sum[block[i]] += abs(color[i] - x);
      color[i] = x;
    }
  }
  for (i = block[a] + 1; i <= block[b] - 1; i++) {
    if (flag[i] != -1) {
      tag[i] += abs(flag[i] - x);
      flag[i] = x;
    } else {
      for (j = (i - 1) * sz + 1; j <= i * sz; j++) {
        val[j] += abs(color[j] - x);
        sum[i] += abs(color[j] - x);
        color[j] = x;
      }
      flag[i] = x;
    }
  }
}
long long query(int a, int b) {
  long long res = 0, i;
  for (i = a; i <= (block[a] * sz > b ? b : block[a] * sz); i++)
    res += val[i] + tag[block[i]];
  if (block[a] != block[b]) {
    for (i = (block[b] - 1) * sz + 1; i <= b; i++)
      res += val[i] + tag[block[i]];
  }
  for (i = block[a] + 1; i <= block[b] - 1; i++) res += sum[i] + tag[i] * sz;
  return res;
}
int main() {
  read(n), read(m), sz = sqrt(n + 0.5);
  for (i = 1; i <= n; i++)
    color[i] = i, block[i] = (i - 1) / sz + 1, flag[block[i]] = -1;
  for (i = 1; i <= m; i++) {
    read(op), read(l), read(r);
    if (op == 1) {
      read(x);
      update(l, r, x);
    } else {
      printf("%lld\n", query(l, r));
    }
  }
  return 0;
}
