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;
long long sum[100005 << 2], col[100005 << 2], add[100005 << 2];
void build(int l, int r, int rt) {
  sum[rt] = 0;
  col[rt] = 0;
  add[rt] = 0;
  if (l == r) {
    col[rt] = l;
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
}
void pushdown(int l, int r, int rt) {
  if (add[rt]) {
    int m = (l + r) >> 1;
    add[rt << 1] += add[rt];
    add[rt << 1 | 1] += add[rt];
    sum[rt << 1] += (m - l + 1) * add[rt];
    sum[rt << 1 | 1] += (r - m) * add[rt];
    add[rt] = 0;
    if (col[rt] != 0) {
      col[rt << 1] = col[rt];
      col[rt << 1 | 1] = col[rt];
    }
  }
}
void pushup(int rt) {
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
  if (col[rt << 1] == col[rt << 1 | 1] && col[rt << 1] != 0)
    col[rt] = col[rt << 1];
  else
    col[rt] = 0;
}
void update(int L, int R, long long val, int l, int r, int rt) {
  if (L <= l && R >= r && col[rt] != 0) {
    add[rt] += abs(val - col[rt]);
    sum[rt] += (long long)(r - l + 1) * abs(val - col[rt]);
    col[rt] = val;
    return;
  }
  int m = (l + r) >> 1;
  pushdown(l, r, rt);
  if (L <= m) update(L, R, val, l, m, rt << 1);
  if (R > m) update(L, R, val, m + 1, r, rt << 1 | 1);
  pushup(rt);
}
long long query(int L, int R, int l, int r, int rt) {
  if (L <= l && R >= r) {
    return sum[rt];
  }
  int m = (l + r) >> 1;
  pushdown(l, r, rt);
  long long ans = 0;
  if (L <= m) ans += query(L, R, l, m, rt << 1);
  if (R > m) ans += query(L, R, m + 1, r, rt << 1 | 1);
  return ans;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  build(1, n, 1);
  long long x, y, z, a;
  while (m--) {
    scanf("%I64d", &a);
    if (a == 1) {
      scanf("%I64d%I64d%I64d", &x, &y, &z);
      update(x, y, z, 1, n, 1);
    } else {
      scanf("%I64d%I64d", &x, &y);
      printf("%I64d\n", query(x, y, 1, n, 1));
    }
  }
  return 0;
}
