#include <bits/stdc++.h>
using namespace std;
const long long INF = 0x3f3f3f3f;
const long long MAXN = 1e5 + 5;
long long n, m;
struct Tree {
  long long l, r;
  long long sum, f, col;
} tree[MAXN * 4];
void update(long long k) {
  if (tree[k * 2].col == tree[k * 2 + 1].col)
    tree[k].col = tree[k * 2].col;
  else
    tree[k].col = 0;
  tree[k].sum = tree[k * 2].sum + tree[k * 2 + 1].sum;
}
void down(long long k) {
  tree[k * 2].col = tree[k].col;
  tree[k * 2 + 1].col = tree[k].col;
  tree[k * 2].f += tree[k].f;
  tree[k * 2 + 1].f += tree[k].f;
  tree[k * 2].sum += tree[k].f * (tree[k * 2].r - tree[k * 2].l + 1);
  tree[k * 2 + 1].sum +=
      tree[k].f * (tree[k * 2 + 1].r - tree[k * 2 + 1].l + 1);
  tree[k].col = tree[k].f = 0;
}
void build(long long k, long long l, long long r) {
  tree[k].l = l;
  tree[k].r = r;
  tree[k].sum = tree[k].col = tree[k].f = 0;
  if (l == r) {
    tree[k].col = l;
    return;
  }
  long long mid = (l + r) / 2;
  build(k * 2, l, mid);
  build(k * 2 + 1, mid + 1, r);
}
void modify(long long k, long long l, long long r, long long val) {
  if (tree[k].l >= l && tree[k].r <= r && tree[k].col) {
    tree[k].sum += (tree[k].r - tree[k].l + 1) * abs(val - tree[k].col);
    tree[k].f += abs(val - tree[k].col);
    tree[k].col = val;
    return;
  }
  if (tree[k].col) down(k);
  long long mid = (tree[k].l + tree[k].r) / 2;
  if (l <= mid) modify(k * 2, l, r, val);
  if (r > mid) modify(k * 2 + 1, l, r, val);
  update(k);
}
long long query(long long k, long long l, long long r) {
  if (tree[k].l >= l && tree[k].r <= r) return tree[k].sum;
  if (tree[k].col) down(k);
  long long mid = (tree[k].l + tree[k].r) / 2;
  long long res = 0;
  if (l <= mid) res += query(k * 2, l, r);
  if (r > mid) res += query(k * 2 + 1, l, r);
  return res;
}
signed main() {
  scanf("%lld%lld", &n, &m);
  build(1, 1, n);
  while (m--) {
    long long op, l, r, x;
    scanf("%lld%lld%lld", &op, &l, &r);
    if (op == 1) {
      scanf("%lld", &x);
      modify(1, l, r, x);
    } else if (op == 2)
      printf("%lld\n", query(1, l, r));
  }
  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 long long INF = 0x3f3f3f3f;
const long long MAXN = 1e5 + 5;
long long n, m;
struct Tree {
  long long l, r;
  long long sum, f, col;
} tree[MAXN * 4];
void update(long long k) {
  if (tree[k * 2].col == tree[k * 2 + 1].col)
    tree[k].col = tree[k * 2].col;
  else
    tree[k].col = 0;
  tree[k].sum = tree[k * 2].sum + tree[k * 2 + 1].sum;
}
void down(long long k) {
  tree[k * 2].col = tree[k].col;
  tree[k * 2 + 1].col = tree[k].col;
  tree[k * 2].f += tree[k].f;
  tree[k * 2 + 1].f += tree[k].f;
  tree[k * 2].sum += tree[k].f * (tree[k * 2].r - tree[k * 2].l + 1);
  tree[k * 2 + 1].sum +=
      tree[k].f * (tree[k * 2 + 1].r - tree[k * 2 + 1].l + 1);
  tree[k].col = tree[k].f = 0;
}
void build(long long k, long long l, long long r) {
  tree[k].l = l;
  tree[k].r = r;
  tree[k].sum = tree[k].col = tree[k].f = 0;
  if (l == r) {
    tree[k].col = l;
    return;
  }
  long long mid = (l + r) / 2;
  build(k * 2, l, mid);
  build(k * 2 + 1, mid + 1, r);
}
void modify(long long k, long long l, long long r, long long val) {
  if (tree[k].l >= l && tree[k].r <= r && tree[k].col) {
    tree[k].sum += (tree[k].r - tree[k].l + 1) * abs(val - tree[k].col);
    tree[k].f += abs(val - tree[k].col);
    tree[k].col = val;
    return;
  }
  if (tree[k].col) down(k);
  long long mid = (tree[k].l + tree[k].r) / 2;
  if (l <= mid) modify(k * 2, l, r, val);
  if (r > mid) modify(k * 2 + 1, l, r, val);
  update(k);
}
long long query(long long k, long long l, long long r) {
  if (tree[k].l >= l && tree[k].r <= r) return tree[k].sum;
  if (tree[k].col) down(k);
  long long mid = (tree[k].l + tree[k].r) / 2;
  long long res = 0;
  if (l <= mid) res += query(k * 2, l, r);
  if (r > mid) res += query(k * 2 + 1, l, r);
  return res;
}
signed main() {
  scanf("%lld%lld", &n, &m);
  build(1, 1, n);
  while (m--) {
    long long op, l, r, x;
    scanf("%lld%lld%lld", &op, &l, &r);
    if (op == 1) {
      scanf("%lld", &x);
      modify(1, l, r, x);
    } else if (op == 2)
      printf("%lld\n", query(1, l, r));
  }
  return 0;
}

```