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;
int n, m;
int abs(int a) { return a >= 0 ? a : -a; }
struct Segtree {
  int color[400010];
  long long delta[400010], sum[400010];
  void update(int x, int l, int r) {
    sum[x] = sum[x << 1] + sum[x << 1 | 1] + 1ll * delta[x] * (r - l + 1);
  }
  void push(int x) {
    if (color[x]) {
      color[x << 1] = color[x << 1 | 1] = color[x];
      color[x] = 0;
    }
  }
  void solve(int x, int l, int r, int c) {
    if (color[x]) {
      delta[x] += abs(c - color[x]);
      sum[x] += 1ll * abs(c - color[x]) * (r - l + 1);
    } else {
      solve((x) << 1, l, (l + r) >> 1, c),
          solve(((x) << 1) | 1, ((l + r) >> 1) + 1, r, c);
      update(x, l, r);
    }
    color[x] = c;
  }
  void paint(int x, int l, int r, int L, int R, int c) {
    if (R < l || r < L) return;
    if (L <= l && r <= R) {
      solve(x, l, r, c);
      return;
    }
    push(x);
    paint((x) << 1, l, (l + r) >> 1, L, R, c),
        paint(((x) << 1) | 1, ((l + r) >> 1) + 1, r, L, R, c);
    update(x, l, r);
  }
  long long query(int x, int l, int r, int L, int R) {
    if (R < l || r < L) return 0;
    if (L <= l && r <= R) return sum[x];
    return query((x) << 1, l, (l + r) >> 1, L, R) +
           query(((x) << 1) | 1, ((l + r) >> 1) + 1, r, L, R) +
           1ll * delta[x] * (min(r, R) - max(l, L) + 1);
  }
  void build(int x, int l, int r) {
    color[x] = delta[x] = sum[x] = 0;
    if (l == r) {
      color[x] = l;
      return;
    }
    build((x) << 1, l, (l + r) >> 1),
        build(((x) << 1) | 1, ((l + r) >> 1) + 1, r);
  }
} seg;
int main() {
  scanf("%d%d", &n, &m);
  seg.build(1, 1, n);
  for (int i = 1, t, l, r, x; i <= m; i++) {
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1) {
      scanf("%d\n", &x);
      seg.paint(1, 1, n, l, r, x);
    } else
      printf("%lld\n", seg.query(1, 1, n, l, r));
  }
  return 0;
}
