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 col[maxn << 2], lazy[maxn << 2], sum[maxn << 2];
void pushup(int cur) {
  sum[cur] = sum[cur << 1] + sum[cur << 1 | 1];
  col[cur] = (col[cur << 1] == col[cur << 1 | 1] ? col[cur << 1] : 0);
}
void build(int l, int r, int cur) {
  if (l == r) {
    col[cur] = l;
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, cur << 1);
  build(m + 1, r, cur << 1 | 1);
  pushup(cur);
}
void pushdown(int cur, int len) {
  if (col[cur]) {
    col[cur << 1] = col[cur << 1 | 1] = col[cur];
    sum[cur << 1] += lazy[cur] * (long long)(len - (len >> 1));
    sum[cur << 1 | 1] += lazy[cur] * (long long)(len >> 1);
    lazy[cur << 1] += lazy[cur];
    lazy[cur << 1 | 1] += lazy[cur];
    col[cur] = lazy[cur] = 0;
  }
}
void updata(int l, int r, int cur, int L, int R, long long val) {
  if (L <= l && r <= R && col[cur]) {
    sum[cur] += abs(col[cur] - val) * (long long)(r - l + 1);
    lazy[cur] += abs(col[cur] - val);
    col[cur] = val;
    return;
  }
  pushdown(cur, r - l + 1);
  int m = (l + r) >> 1;
  if (L <= m) updata(l, m, cur << 1, L, R, val);
  if (R > m) updata(m + 1, r, cur << 1 | 1, L, R, val);
  pushup(cur);
}
long long query(int l, int r, int cur, int L, int R) {
  if (L <= l && r <= R) {
    return sum[cur];
  }
  pushdown(cur, r - l + 1);
  long long ans = 0;
  int m = (l + r) >> 1;
  if (L <= m) ans += query(l, m, cur << 1, L, R);
  if (R > m) ans += query(m + 1, r, cur << 1 | 1, L, R);
  return ans;
}
int main() {
  int n, m, fg, l, r;
  long long x;
  scanf("%d %d", &n, &m);
  build(1, n, 1);
  while (m--) {
    scanf("%d", &fg);
    if (fg == 1) {
      scanf("%d %d %lld", &l, &r, &x);
      updata(1, n, 1, l, r, x);
    } else {
      scanf("%d %d", &l, &r);
      printf("%lld\n", query(1, n, 1, l, r));
    }
  }
}
