#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10;
long long col[maxn << 2], add[maxn << 2], sum[maxn << 2];
int op, qL, qR, v;
void build(int o, int L, int R) {
  if (L == R) {
    col[o] = L;
    return;
  }
  int M = (L + R) / 2;
  build(o << 1, L, M);
  build(o << 1 | 1, M + 1, R);
}
void Paint(int o, int L, int R) {
  if (col[o]) {
    add[o] += (long long)abs(col[o] - v);
    sum[o] += (long long)abs(col[o] - v) * (R - L + 1);
  } else {
    int M = (L + R) / 2;
    Paint(o << 1, L, M);
    Paint(o << 1 | 1, M + 1, R);
    sum[o] = sum[o << 1] + sum[o << 1 | 1] + (long long)add[o] * (R - L + 1);
  }
  col[o] = 0;
}
void update(int o, int L, int R) {
  if (qL <= L && qR >= R) {
    Paint(o, L, R);
    col[o] = v;
    return;
  }
  if (col[o]) {
    col[o << 1] = col[o << 1 | 1] = col[o];
    col[o] = 0;
  }
  int M = (L + R) / 2;
  if (qL <= M) update(o << 1, L, M);
  if (qR > M) update(o << 1 | 1, M + 1, R);
  col[o] = 0;
  sum[o] = sum[o << 1] + sum[o << 1 | 1] + (long long)add[o] * (R - L + 1);
}
long long query(int o, int L, int R, int qL, int qR) {
  if (qL <= L && qR >= R) return sum[o];
  int M = (L + R) / 2;
  if (qR <= M)
    return query(o << 1, L, M, qL, qR) + (long long)add[o] * (qR - qL + 1);
  if (qL > M)
    return query(o << 1 | 1, M + 1, R, qL, qR) +
           (long long)add[o] * (qR - qL + 1);
  return query(o << 1, L, M, qL, M) + query(o << 1 | 1, M + 1, R, M + 1, qR) +
         (long long)add[o] * (qR - qL + 1);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  while (m--) {
    scanf("%d%d%d", &op, &qL, &qR);
    if (op == 1) {
      scanf("%d", &v);
      update(1, 1, n);
    } else
      printf("%I64d\n", query(1, 1, n, qL, qR));
  }
  return 0;
}

### Prompt

Generate a Cpp solution to 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 = 100000 + 10;
long long col[maxn << 2], add[maxn << 2], sum[maxn << 2];
int op, qL, qR, v;
void build(int o, int L, int R) {
  if (L == R) {
    col[o] = L;
    return;
  }
  int M = (L + R) / 2;
  build(o << 1, L, M);
  build(o << 1 | 1, M + 1, R);
}
void Paint(int o, int L, int R) {
  if (col[o]) {
    add[o] += (long long)abs(col[o] - v);
    sum[o] += (long long)abs(col[o] - v) * (R - L + 1);
  } else {
    int M = (L + R) / 2;
    Paint(o << 1, L, M);
    Paint(o << 1 | 1, M + 1, R);
    sum[o] = sum[o << 1] + sum[o << 1 | 1] + (long long)add[o] * (R - L + 1);
  }
  col[o] = 0;
}
void update(int o, int L, int R) {
  if (qL <= L && qR >= R) {
    Paint(o, L, R);
    col[o] = v;
    return;
  }
  if (col[o]) {
    col[o << 1] = col[o << 1 | 1] = col[o];
    col[o] = 0;
  }
  int M = (L + R) / 2;
  if (qL <= M) update(o << 1, L, M);
  if (qR > M) update(o << 1 | 1, M + 1, R);
  col[o] = 0;
  sum[o] = sum[o << 1] + sum[o << 1 | 1] + (long long)add[o] * (R - L + 1);
}
long long query(int o, int L, int R, int qL, int qR) {
  if (qL <= L && qR >= R) return sum[o];
  int M = (L + R) / 2;
  if (qR <= M)
    return query(o << 1, L, M, qL, qR) + (long long)add[o] * (qR - qL + 1);
  if (qL > M)
    return query(o << 1 | 1, M + 1, R, qL, qR) +
           (long long)add[o] * (qR - qL + 1);
  return query(o << 1, L, M, qL, M) + query(o << 1 | 1, M + 1, R, M + 1, qR) +
         (long long)add[o] * (qR - qL + 1);
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  while (m--) {
    scanf("%d%d%d", &op, &qL, &qR);
    if (op == 1) {
      scanf("%d", &v);
      update(1, 1, n);
    } else
      printf("%I64d\n", query(1, 1, n, qL, qR));
  }
  return 0;
}

```