#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 1;
long long tr[maxn << 2], col[maxn << 2], lazy[maxn << 2];
void build(int id, int l, int r) {
  if (l == r) {
    tr[id] = l * 1LL;
    col[id] = lazy[id] = 0;
  } else {
    int mid = (l + r) >> 1;
    build((id << 1), l, mid);
    build(((id << 1) | 1), mid + 1, r);
    tr[id] = col[id] = lazy[id] = 0;
  }
}
void push_down(int id, int l, int r) {
  int mid = (l + r) >> 1;
  col[(id << 1)] += lazy[id] * (mid - l + 1LL);
  col[((id << 1) | 1)] += lazy[id] * (r - mid * 1LL);
  lazy[(id << 1)] += lazy[id];
  lazy[((id << 1) | 1)] += lazy[id];
  tr[(id << 1)] = tr[((id << 1) | 1)] = tr[id];
  lazy[id] = 0;
}
void update(int id, int l, int r, int L, int R, long long x) {
  if (L <= l && r <= R && tr[id]) {
    col[id] += (abs(tr[id] - x) * (r - l + 1LL));
    lazy[id] += (abs(tr[id] - x));
    tr[id] = x;
  } else {
    if (lazy[id]) push_down(id, l, r);
    int mid = (l + r) >> 1;
    if (L <= l && r <= R)
      tr[id] = x;
    else
      tr[id] = 0;
    if (mid >= L) update((id << 1), l, mid, L, R, x);
    if (mid < R) update(((id << 1) | 1), mid + 1, r, L, R, x);
    col[id] = col[(id << 1)] + col[((id << 1) | 1)];
  }
}
long long query(int id, int l, int r, int L, int R) {
  if (L <= l && r <= R)
    return col[id];
  else {
    if (lazy[id]) push_down(id, l, r);
    int mid = (l + r) >> 1;
    long long res = 0;
    if (mid >= L) res += query((id << 1), l, mid, L, R);
    if (mid < R) res += query(((id << 1) | 1), mid + 1, r, L, R);
    return res;
  }
}
int main() {
  int n, m, cmd, i, l, r;
  long long x;
  scanf("%d %d", &n, &m);
  build(1, 1, n);
  for (i = 0; i < m; i++) {
    scanf("%d", &cmd);
    if (cmd == 1) {
      scanf("%d %d %I64d", &l, &r, &x);
      update(1, 1, n, l, r, x);
    } else {
      scanf("%d %d", &l, &r);
      printf("%I64d\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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 + 1;
long long tr[maxn << 2], col[maxn << 2], lazy[maxn << 2];
void build(int id, int l, int r) {
  if (l == r) {
    tr[id] = l * 1LL;
    col[id] = lazy[id] = 0;
  } else {
    int mid = (l + r) >> 1;
    build((id << 1), l, mid);
    build(((id << 1) | 1), mid + 1, r);
    tr[id] = col[id] = lazy[id] = 0;
  }
}
void push_down(int id, int l, int r) {
  int mid = (l + r) >> 1;
  col[(id << 1)] += lazy[id] * (mid - l + 1LL);
  col[((id << 1) | 1)] += lazy[id] * (r - mid * 1LL);
  lazy[(id << 1)] += lazy[id];
  lazy[((id << 1) | 1)] += lazy[id];
  tr[(id << 1)] = tr[((id << 1) | 1)] = tr[id];
  lazy[id] = 0;
}
void update(int id, int l, int r, int L, int R, long long x) {
  if (L <= l && r <= R && tr[id]) {
    col[id] += (abs(tr[id] - x) * (r - l + 1LL));
    lazy[id] += (abs(tr[id] - x));
    tr[id] = x;
  } else {
    if (lazy[id]) push_down(id, l, r);
    int mid = (l + r) >> 1;
    if (L <= l && r <= R)
      tr[id] = x;
    else
      tr[id] = 0;
    if (mid >= L) update((id << 1), l, mid, L, R, x);
    if (mid < R) update(((id << 1) | 1), mid + 1, r, L, R, x);
    col[id] = col[(id << 1)] + col[((id << 1) | 1)];
  }
}
long long query(int id, int l, int r, int L, int R) {
  if (L <= l && r <= R)
    return col[id];
  else {
    if (lazy[id]) push_down(id, l, r);
    int mid = (l + r) >> 1;
    long long res = 0;
    if (mid >= L) res += query((id << 1), l, mid, L, R);
    if (mid < R) res += query(((id << 1) | 1), mid + 1, r, L, R);
    return res;
  }
}
int main() {
  int n, m, cmd, i, l, r;
  long long x;
  scanf("%d %d", &n, &m);
  build(1, 1, n);
  for (i = 0; i < m; i++) {
    scanf("%d", &cmd);
    if (cmd == 1) {
      scanf("%d %d %I64d", &l, &r, &x);
      update(1, 1, n, l, r, x);
    } else {
      scanf("%d %d", &l, &r);
      printf("%I64d\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}

```