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 N = 100009;
const long long oo = 1000;
int n, query, u, v, val, type;
long long zip[N * 4], zipsum[N * 4], IT[N * 4];
void build(int i, int l, int r) {
  if (l > r) return;
  if (l == r) {
    zip[i] = l;
    return;
  }
  int mid = (l + r) >> 1;
  build(i << 1, l, mid);
  build((i << 1) + 1, mid + 1, r);
}
void update(int i, int l, int r, int u, int v, int val) {
  if (l > r || l > v || r < u) return;
  if (l >= u && r <= v && zip[i]) {
    zipsum[i] += abs(val - zip[i]);
    zip[i] = val;
    return;
  }
  int mid = (l + r) >> 1;
  int i1 = i << 1, i2 = i1 + 1;
  if (zip[i]) {
    zipsum[i1] += zipsum[i];
    zipsum[i2] += zipsum[i];
    zip[i1] = zip[i2] = zip[i];
    zipsum[i] = zip[i] = 0;
  }
  update(i1, l, mid, u, v, val);
  update(i2, mid + 1, r, u, v, val);
  IT[i] = IT[i1] + IT[i2] + zipsum[i1] * (mid - l + 1) + zipsum[i2] * (r - mid);
  if (zip[i1] && zip[i2] && (zip[i1] == zip[i2])) zip[i] = zip[i1];
}
long long get(int i, int l, int r, int u, int v) {
  if (l > r || l > v || r < u) return 0ll;
  if (l >= u && r <= v) return (IT[i] + zipsum[i] * (r - l + 1));
  int mid = (l + r) >> 1;
  int i1 = i << 1, i2 = i1 + 1;
  if (zip[i]) {
    zipsum[i1] += zipsum[i];
    zipsum[i2] += zipsum[i];
    zip[i1] = zip[i2] = zip[i];
    zipsum[i] = zip[i] = 0;
  }
  long long ret = get(i1, l, mid, u, v) + get(i2, mid + 1, r, u, v);
  IT[i] = IT[i1] + IT[i2] + zipsum[i1] * (mid - l + 1) + zipsum[i2] * (r - mid);
  if (zip[i1] && zip[i2] && (zip[i1] == zip[i2])) zip[i] = zip[i1];
  return ret;
}
int main() {
  scanf("%d %d", &n, &query);
  build(1, 1, n);
  while (query--) {
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d %d %d", &u, &v, &val);
      update(1, 1, n, u, v, val);
    } else {
      scanf("%d %d", &u, &v);
      long long ans = get(1, 1, n, u, v);
      printf("%I64d\n", ans);
    }
  }
}
