#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * power(a * a % md, b / 2, md) % md
                      : power(a * a % md, b / 2, md) % md));
}
const int xn = 1e5 + 10;
const int xm = 20 + 10;
const int SQ = 320;
const int sq = 500 + 10;
const int inf = 1e9 + 10;
const long long INF = 1e14 + 10;
const int MOD = 1e9 + 7;
const int TM = 1e7;
const int BASE = 257;
int n, m, ptr, lazy[xn << 2];
long long seg[xn << 2], val[xn << 2], sum[xn << 2];
void build(int id, int l, int r) {
  if (r - l == 1) {
    val[id] = l;
    return;
  }
  int mid = l + r >> 1;
  build(id << 1, l, mid);
  build(id << 1 | 1, mid, r);
}
void shift(int id, int l, int r) {
  if (!lazy[id]) return;
  val[id] = lazy[id];
  seg[id] += sum[id] * (r - l);
  if (r - l > 1) {
    lazy[id << 1] = lazy[id];
    lazy[id << 1 | 1] = lazy[id];
    sum[id << 1] += sum[id];
    sum[id << 1 | 1] += sum[id];
  }
  sum[id] = lazy[id] = 0;
}
void update(int id, int l, int r, int ql, int qr, long long x) {
  shift(id, l, r);
  if (qr <= l || r <= ql) return;
  if (ql <= l && r <= qr && val[id]) {
    lazy[id] = x;
    sum[id] += abs(x - val[id]);
    shift(id, l, r);
    return;
  }
  int mid = l + r >> 1;
  update(id << 1, l, mid, ql, qr, x);
  update(id << 1 | 1, mid, r, ql, qr, x);
  seg[id] = seg[id << 1] + seg[id << 1 | 1];
  if (val[id << 1] == val[id << 1 | 1])
    val[id] = val[id << 1];
  else
    val[id] = 0;
}
long long get(int id, int l, int r, int ql, int qr) {
  shift(id, l, r);
  if (qr <= l || r <= ql) return 0;
  if (ql <= l && r <= qr) return seg[id];
  int mid = l + r >> 1;
  return get(id << 1, l, mid, ql, qr) + get(id << 1 | 1, mid, r, ql, qr);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  build(1, 1, n + 1);
  while (m--) {
    int t, l, r, x;
    cin >> t >> l >> r;
    if (t == 1) {
      cin >> x;
      update(1, 1, n + 1, l, r + 1, x);
    } else
      cout << get(1, 1, n + 1, l, r + 1) << '\n';
  }
  return 0;
}

### Prompt

Your task is to create 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;
long long power(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * power(a * a % md, b / 2, md) % md
                      : power(a * a % md, b / 2, md) % md));
}
const int xn = 1e5 + 10;
const int xm = 20 + 10;
const int SQ = 320;
const int sq = 500 + 10;
const int inf = 1e9 + 10;
const long long INF = 1e14 + 10;
const int MOD = 1e9 + 7;
const int TM = 1e7;
const int BASE = 257;
int n, m, ptr, lazy[xn << 2];
long long seg[xn << 2], val[xn << 2], sum[xn << 2];
void build(int id, int l, int r) {
  if (r - l == 1) {
    val[id] = l;
    return;
  }
  int mid = l + r >> 1;
  build(id << 1, l, mid);
  build(id << 1 | 1, mid, r);
}
void shift(int id, int l, int r) {
  if (!lazy[id]) return;
  val[id] = lazy[id];
  seg[id] += sum[id] * (r - l);
  if (r - l > 1) {
    lazy[id << 1] = lazy[id];
    lazy[id << 1 | 1] = lazy[id];
    sum[id << 1] += sum[id];
    sum[id << 1 | 1] += sum[id];
  }
  sum[id] = lazy[id] = 0;
}
void update(int id, int l, int r, int ql, int qr, long long x) {
  shift(id, l, r);
  if (qr <= l || r <= ql) return;
  if (ql <= l && r <= qr && val[id]) {
    lazy[id] = x;
    sum[id] += abs(x - val[id]);
    shift(id, l, r);
    return;
  }
  int mid = l + r >> 1;
  update(id << 1, l, mid, ql, qr, x);
  update(id << 1 | 1, mid, r, ql, qr, x);
  seg[id] = seg[id << 1] + seg[id << 1 | 1];
  if (val[id << 1] == val[id << 1 | 1])
    val[id] = val[id << 1];
  else
    val[id] = 0;
}
long long get(int id, int l, int r, int ql, int qr) {
  shift(id, l, r);
  if (qr <= l || r <= ql) return 0;
  if (ql <= l && r <= qr) return seg[id];
  int mid = l + r >> 1;
  return get(id << 1, l, mid, ql, qr) + get(id << 1 | 1, mid, r, ql, qr);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  build(1, 1, n + 1);
  while (m--) {
    int t, l, r, x;
    cin >> t >> l >> r;
    if (t == 1) {
      cin >> x;
      update(1, 1, n + 1, l, r + 1, x);
    } else
      cout << get(1, 1, n + 1, l, r + 1) << '\n';
  }
  return 0;
}

```