#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
long long col[MAXN << 2], chg[MAXN << 2], sum[MAXN << 2];
void push_down(int n, int len) {
  if (col[n]) {
    chg[n << 1] += chg[n];
    chg[n << 1 | 1] += chg[n];
    sum[n << 1] += chg[n] * (len - (len >> 1));
    sum[n << 1 | 1] += chg[n] * (len >> 1);
    col[n << 1] = col[n];
    col[n << 1 | 1] = col[n];
    col[n] = chg[n] = 0;
  }
}
void push_up(int n) {
  if (col[n << 1] == col[n << 1 | 1]) col[n] = col[n << 1];
  sum[n] = sum[n << 1] + sum[n << 1 | 1];
}
void build(int L, int R, int n) {
  sum[n] = 0;
  if (L == R) {
    col[n] = L;
    return;
  }
  int mid = (L + R) >> 1;
  build(L, mid, n << 1);
  build(mid + 1, R, n << 1 | 1);
}
void modify(int L, int R, int n, int ls, int rs, int val) {
  if (ls <= L && R <= rs && col[n]) {
    chg[n] += abs(col[n] - val);
    sum[n] += abs(col[n] - val) * (R - L + 1);
    col[n] = val;
    return;
  }
  push_down(n, R - L + 1);
  int mid = (L + R) >> 1;
  if (ls <= mid) modify(L, mid, n << 1, ls, rs, val);
  if (rs > mid) modify(mid + 1, R, n << 1 | 1, ls, rs, val);
  push_up(n);
}
long long query(int L, int R, int n, int ls, int rs) {
  if (ls <= L && R <= rs) {
    return sum[n];
  }
  push_down(n, R - L + 1);
  int mid = (L + R) >> 1;
  long long ret = 0;
  if (ls <= mid) ret += query(L, mid, n << 1, ls, rs);
  if (rs > mid) ret += query(mid + 1, R, n << 1 | 1, ls, rs);
  return ret;
}
int main() {
  int n, m;
  int t, l, r, x;
  scanf("%d%d", &n, &m);
  build(1, n, 1);
  while (m--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d%d", &l, &r, &x);
      modify(1, n, 1, l, r, x);
    } else {
      scanf("%d%d", &l, &r);
      printf("%lld\n", query(1, n, 1, l, r));
    }
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
long long col[MAXN << 2], chg[MAXN << 2], sum[MAXN << 2];
void push_down(int n, int len) {
  if (col[n]) {
    chg[n << 1] += chg[n];
    chg[n << 1 | 1] += chg[n];
    sum[n << 1] += chg[n] * (len - (len >> 1));
    sum[n << 1 | 1] += chg[n] * (len >> 1);
    col[n << 1] = col[n];
    col[n << 1 | 1] = col[n];
    col[n] = chg[n] = 0;
  }
}
void push_up(int n) {
  if (col[n << 1] == col[n << 1 | 1]) col[n] = col[n << 1];
  sum[n] = sum[n << 1] + sum[n << 1 | 1];
}
void build(int L, int R, int n) {
  sum[n] = 0;
  if (L == R) {
    col[n] = L;
    return;
  }
  int mid = (L + R) >> 1;
  build(L, mid, n << 1);
  build(mid + 1, R, n << 1 | 1);
}
void modify(int L, int R, int n, int ls, int rs, int val) {
  if (ls <= L && R <= rs && col[n]) {
    chg[n] += abs(col[n] - val);
    sum[n] += abs(col[n] - val) * (R - L + 1);
    col[n] = val;
    return;
  }
  push_down(n, R - L + 1);
  int mid = (L + R) >> 1;
  if (ls <= mid) modify(L, mid, n << 1, ls, rs, val);
  if (rs > mid) modify(mid + 1, R, n << 1 | 1, ls, rs, val);
  push_up(n);
}
long long query(int L, int R, int n, int ls, int rs) {
  if (ls <= L && R <= rs) {
    return sum[n];
  }
  push_down(n, R - L + 1);
  int mid = (L + R) >> 1;
  long long ret = 0;
  if (ls <= mid) ret += query(L, mid, n << 1, ls, rs);
  if (rs > mid) ret += query(mid + 1, R, n << 1 | 1, ls, rs);
  return ret;
}
int main() {
  int n, m;
  int t, l, r, x;
  scanf("%d%d", &n, &m);
  build(1, n, 1);
  while (m--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d%d", &l, &r, &x);
      modify(1, n, 1, l, r, x);
    } else {
      scanf("%d%d", &l, &r);
      printf("%lld\n", query(1, n, 1, l, r));
    }
  }
  return 0;
}

```