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 = 1e5 + 10;
int n, m;
struct node {
  int l, r;
  long long col, sum, lazy;
} t[maxn << 2];
void pushup(int k) {
  if (t[k << 1].col == t[k << 1 | 1].col) {
    t[k].col = t[k << 1].col;
  } else {
    t[k].col = -1;
  }
  t[k].sum = t[k << 1].sum + t[k << 1 | 1].sum;
}
void build(int k, int l, int r) {
  t[k].l = l, t[k].r = r, t[k].lazy = 0;
  if (l == r) {
    t[k].col = l;
    t[k].sum = 0;
  } else {
    int mid = (l + r) >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
    pushup(k);
  }
}
void pushdown(int k) {
  if (t[k].lazy) {
    t[k << 1].sum += t[k].lazy * (t[k << 1].r - t[k << 1].l + 1);
    t[k << 1 | 1].sum += t[k].lazy * (t[k << 1 | 1].r - t[k << 1 | 1].l + 1);
    t[k << 1].col = t[k << 1 | 1].col = t[k].col;
    t[k << 1].lazy += t[k].lazy;
    t[k << 1 | 1].lazy += t[k].lazy;
    t[k].lazy = 0;
  }
}
void updata(int k, int l, int r, long long v) {
  if (l <= t[k].l && t[k].r <= r) {
    if (t[k].col != -1) {
      t[k].sum += ((long long)(abs(t[k].col - v)) * (t[k].r - t[k].l + 1));
      t[k].lazy += (long long)abs(t[k].col - v);
      t[k].col = v;
      return;
    }
  }
  pushdown(k);
  int mid = (t[k].l + t[k].r) >> 1;
  if (l <= mid) updata(k << 1, l, r, v);
  if (mid < r) updata(k << 1 | 1, l, r, v);
  pushup(k);
}
long long query(int k, int l, int r) {
  if (l <= t[k].l && t[k].r <= r) {
    return t[k].sum;
  } else {
    pushdown(k);
    int mid = (t[k].l + t[k].r) >> 1;
    long long ans = 0;
    if (l <= mid) ans += query(k << 1, l, r);
    if (mid < r) ans += query(k << 1 | 1, l, r);
    return ans;
  }
}
int main() {
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  while (m--) {
    int op, l, r;
    scanf("%d", &op);
    if (op == 1) {
      long long x;
      scanf("%d%d%lld", &l, &r, &x);
      updata(1, l, r, x);
    }
    if (op == 2) {
      scanf("%d%d", &l, &r);
      printf("%lld\n", query(1, l, r));
    }
  }
  return 0;
}
