#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
struct node {
  int l, r, col;
  long long lazy, sum;
} tr[maxn << 2];
void pushup(int m) {
  tr[m].sum = tr[m << 1].sum + tr[m << 1 | 1].sum;
  tr[m].col = tr[m << 1].col == tr[m << 1 | 1].col ? tr[m << 1].col : 0;
}
void pushdown(int m) {
  if (tr[m].col) {
    tr[m << 1].col = tr[m << 1 | 1].col = tr[m].col;
    tr[m << 1].sum += tr[m].lazy * (long long)(tr[m << 1].r - tr[m << 1].l + 1);
    tr[m << 1 | 1].sum +=
        tr[m].lazy * (long long)(tr[m << 1 | 1].r - tr[m << 1 | 1].l + 1);
    tr[m << 1].lazy += tr[m].lazy;
    tr[m << 1 | 1].lazy += tr[m].lazy;
    tr[m].lazy = 0;
  }
}
void build(int m, int l, int r) {
  tr[m].l = l;
  tr[m].r = r;
  tr[m].lazy = 0;
  if (l == r) {
    tr[m].col = l;
    tr[m].sum = 0;
    return;
  }
  int mid = (l + r) >> 1;
  build(m << 1, l, mid);
  build(m << 1 | 1, mid + 1, r);
  pushup(m);
}
void updata(int m, int l, int r, int val) {
  if (tr[m].col && tr[m].l == l && tr[m].r == r) {
    tr[m].sum +=
        (long long)(tr[m].r - tr[m].l + 1) * (long long)abs(val - tr[m].col);
    tr[m].lazy += (long long)abs(val - tr[m].col);
    tr[m].col = val;
    return;
  }
  pushdown(m);
  int mid = (tr[m].l + tr[m].r) >> 1;
  if (r <= mid)
    updata(m << 1, l, r, val);
  else if (l > mid)
    updata(m << 1 | 1, l, r, val);
  else {
    updata(m << 1, l, mid, val);
    updata(m << 1 | 1, mid + 1, r, val);
  }
  pushup(m);
}
long long query(int m, int l, int r) {
  if (tr[m].l == l && tr[m].r == r) {
    return tr[m].sum;
  }
  pushdown(m);
  int mid = (tr[m].l + tr[m].r) >> 1;
  long long temp;
  if (r <= mid)
    temp = query(m << 1, l, r);
  else if (l > mid)
    temp = query(m << 1 | 1, l, r);
  else {
    temp = query(m << 1, l, mid) + query(m << 1 | 1, mid + 1, r);
  }
  pushup(m);
  return temp;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  while (m--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int l, r, x;
      scanf("%d%d%d", &l, &r, &x);
      updata(1, l, r, x);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%lld\n", query(1, l, r));
    }
  }
  return 0;
}

### Prompt

Please formulate 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 = 1e5 + 10;
struct node {
  int l, r, col;
  long long lazy, sum;
} tr[maxn << 2];
void pushup(int m) {
  tr[m].sum = tr[m << 1].sum + tr[m << 1 | 1].sum;
  tr[m].col = tr[m << 1].col == tr[m << 1 | 1].col ? tr[m << 1].col : 0;
}
void pushdown(int m) {
  if (tr[m].col) {
    tr[m << 1].col = tr[m << 1 | 1].col = tr[m].col;
    tr[m << 1].sum += tr[m].lazy * (long long)(tr[m << 1].r - tr[m << 1].l + 1);
    tr[m << 1 | 1].sum +=
        tr[m].lazy * (long long)(tr[m << 1 | 1].r - tr[m << 1 | 1].l + 1);
    tr[m << 1].lazy += tr[m].lazy;
    tr[m << 1 | 1].lazy += tr[m].lazy;
    tr[m].lazy = 0;
  }
}
void build(int m, int l, int r) {
  tr[m].l = l;
  tr[m].r = r;
  tr[m].lazy = 0;
  if (l == r) {
    tr[m].col = l;
    tr[m].sum = 0;
    return;
  }
  int mid = (l + r) >> 1;
  build(m << 1, l, mid);
  build(m << 1 | 1, mid + 1, r);
  pushup(m);
}
void updata(int m, int l, int r, int val) {
  if (tr[m].col && tr[m].l == l && tr[m].r == r) {
    tr[m].sum +=
        (long long)(tr[m].r - tr[m].l + 1) * (long long)abs(val - tr[m].col);
    tr[m].lazy += (long long)abs(val - tr[m].col);
    tr[m].col = val;
    return;
  }
  pushdown(m);
  int mid = (tr[m].l + tr[m].r) >> 1;
  if (r <= mid)
    updata(m << 1, l, r, val);
  else if (l > mid)
    updata(m << 1 | 1, l, r, val);
  else {
    updata(m << 1, l, mid, val);
    updata(m << 1 | 1, mid + 1, r, val);
  }
  pushup(m);
}
long long query(int m, int l, int r) {
  if (tr[m].l == l && tr[m].r == r) {
    return tr[m].sum;
  }
  pushdown(m);
  int mid = (tr[m].l + tr[m].r) >> 1;
  long long temp;
  if (r <= mid)
    temp = query(m << 1, l, r);
  else if (l > mid)
    temp = query(m << 1 | 1, l, r);
  else {
    temp = query(m << 1, l, mid) + query(m << 1 | 1, mid + 1, r);
  }
  pushup(m);
  return temp;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  while (m--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int l, r, x;
      scanf("%d%d%d", &l, &r, &x);
      updata(1, l, r, x);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%lld\n", query(1, l, r));
    }
  }
  return 0;
}

```