#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
long long n, m, k, l, r;
long long x;
struct SegTree {
  long long l, r;
  long long data, sum;
  long long lazy;
} t[4 * N];
void build(long long p, long long l, long long r) {
  t[p].l = l;
  t[p].r = r;
  if (l == r) {
    t[p].data = l;
    return;
  }
  long long mid = l + r >> 1;
  build(p * 2, l, mid);
  build(p * 2 + 1, mid + 1, r);
}
void spread(long long p) {
  if (t[p].lazy) {
    t[p * 2].data = t[p * 2 + 1].data = t[p].data;
    t[p * 2].lazy += t[p].lazy;
    t[p * 2 + 1].lazy += t[p].lazy;
    t[p * 2].sum += (t[p * 2].r - t[p * 2].l + 1) * t[p].lazy;
    t[p * 2 + 1].sum += (t[p * 2 + 1].r - t[p * 2 + 1].l + 1) * t[p].lazy;
    t[p].lazy = 0;
    t[p].data = 0;
  }
}
void change(long long p, long long l, long long r, long long x) {
  if (t[p].l >= l && t[p].r <= r && t[p].data) {
    t[p].sum += abs(t[p].data - x) * (t[p].r - t[p].l + 1);
    t[p].lazy += abs(t[p].data - x);
    t[p].data = x;
  } else {
    spread(p);
    long long mid = t[p].l + t[p].r >> 1;
    if (mid >= l) change(p * 2, l, r, x);
    if (mid < r) change(p * 2 + 1, l, r, x);
    t[p].sum = t[p * 2].sum + t[p * 2 + 1].sum;
    t[p].data = t[p * 2].data == t[p * 2 + 1].data ? t[p * 2].data : 0;
  }
}
long long ask(long long p, long long l, long long r) {
  if (t[p].l >= l && t[p].r <= r) {
    return t[p].sum;
  }
  spread(p);
  long long mid = t[p].l + t[p].r >> 1;
  long long res = 0;
  if (mid >= l) res += ask(p * 2, l, r);
  if (mid < r) res += ask(p * 2 + 1, l, r);
  return res;
}
int main() {
  scanf("%lld%lld", &n, &m);
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    scanf("%lld%lld%lld", &k, &l, &r);
    if (k == 1) {
      scanf("%lld", &x);
      change(1, l, r, x);
    } else {
      printf("%lld\n", ask(1, l, r));
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 N = 100010;
long long n, m, k, l, r;
long long x;
struct SegTree {
  long long l, r;
  long long data, sum;
  long long lazy;
} t[4 * N];
void build(long long p, long long l, long long r) {
  t[p].l = l;
  t[p].r = r;
  if (l == r) {
    t[p].data = l;
    return;
  }
  long long mid = l + r >> 1;
  build(p * 2, l, mid);
  build(p * 2 + 1, mid + 1, r);
}
void spread(long long p) {
  if (t[p].lazy) {
    t[p * 2].data = t[p * 2 + 1].data = t[p].data;
    t[p * 2].lazy += t[p].lazy;
    t[p * 2 + 1].lazy += t[p].lazy;
    t[p * 2].sum += (t[p * 2].r - t[p * 2].l + 1) * t[p].lazy;
    t[p * 2 + 1].sum += (t[p * 2 + 1].r - t[p * 2 + 1].l + 1) * t[p].lazy;
    t[p].lazy = 0;
    t[p].data = 0;
  }
}
void change(long long p, long long l, long long r, long long x) {
  if (t[p].l >= l && t[p].r <= r && t[p].data) {
    t[p].sum += abs(t[p].data - x) * (t[p].r - t[p].l + 1);
    t[p].lazy += abs(t[p].data - x);
    t[p].data = x;
  } else {
    spread(p);
    long long mid = t[p].l + t[p].r >> 1;
    if (mid >= l) change(p * 2, l, r, x);
    if (mid < r) change(p * 2 + 1, l, r, x);
    t[p].sum = t[p * 2].sum + t[p * 2 + 1].sum;
    t[p].data = t[p * 2].data == t[p * 2 + 1].data ? t[p * 2].data : 0;
  }
}
long long ask(long long p, long long l, long long r) {
  if (t[p].l >= l && t[p].r <= r) {
    return t[p].sum;
  }
  spread(p);
  long long mid = t[p].l + t[p].r >> 1;
  long long res = 0;
  if (mid >= l) res += ask(p * 2, l, r);
  if (mid < r) res += ask(p * 2 + 1, l, r);
  return res;
}
int main() {
  scanf("%lld%lld", &n, &m);
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    scanf("%lld%lld%lld", &k, &l, &r);
    if (k == 1) {
      scanf("%lld", &x);
      change(1, l, r, x);
    } else {
      printf("%lld\n", ask(1, l, r));
    }
  }
  return 0;
}

```