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 = 100010;
struct tree {
  int l, r;
  long long color, sum, add;
} a[MAXN * 4];
int n, m;
void push_up(int k) {
  a[k].sum = a[k << 1].sum + a[k << 1 | 1].sum;
  if (a[k << 1].color > 0 && a[k << 1].color == a[k << 1 | 1].color)
    a[k].color = a[k << 1].color;
  else
    a[k].color = 0;
}
void push_down(int k) {
  if (a[k].add > 0) {
    a[k << 1].add += a[k].add;
    a[k << 1 | 1].add += a[k].add;
    a[k << 1].color = a[k << 1 | 1].color = a[k].color;
    a[k << 1].sum += a[k].add * (a[k << 1].r - a[k << 1].l + 1);
    a[k << 1 | 1].sum += a[k].add * (a[k << 1 | 1].r - a[k << 1 | 1].l + 1);
    a[k].add = 0;
  }
}
void build(int k, int l, int r) {
  a[k].l = l;
  a[k].r = r;
  a[k].sum = a[k].add = 0;
  if (l == r) {
    a[k].color = l;
  } else
    a[k].color = 0;
  if (l == r) return;
  int mid = (l + r) / 2;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
  push_up(k);
}
void add(int k, int l, int r, long long color) {
  if (l <= a[k].l && a[k].r <= r && a[k].color > 0) {
    a[k].add += abs(color - a[k].color);
    int len = a[k].r - a[k].l + 1;
    a[k].sum += abs(color - a[k].color) * len;
    a[k].color = color;
    return;
  }
  push_down(k);
  int mid = (a[k].l + a[k].r) / 2;
  if (mid >= r)
    add(k << 1, l, r, color);
  else if (mid < l)
    add(k << 1 | 1, l, r, color);
  else {
    add(k << 1, l, mid, color);
    add(k << 1 | 1, mid + 1, r, color);
  }
  push_up(k);
}
long long query(int k, int l, int r) {
  if (l <= a[k].l && a[k].r <= r) {
    return a[k].sum;
  }
  push_down(k);
  int mid = (a[k].l + a[k].r) / 2;
  if (mid >= r)
    return query(k << 1, l, r);
  else if (mid < l)
    return query(k << 1 | 1, l, r);
  else {
    long long tmp = 0;
    tmp += query(k << 1, l, mid);
    tmp += query(k << 1 | 1, mid + 1, r);
    return tmp;
  }
}
int main() {
  cin >> n >> m;
  build(1, 1, n);
  for (int i = 0; i < m; i++) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int l, r;
      long long x;
      scanf("%d%d%I64d", &l, &r, &x);
      add(1, l, r, x);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%I64d\n", query(1, l, r));
    }
  }
  return 0;
}
