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.
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100007;
struct nd {
  long long same, lazy;
  long long sum;
} tr[MAXN * 4];
void init(int u, int l, int r) {
  tr[u].same = tr[u].sum = 0;
  tr[u].lazy = 0;
  if (l == r) {
    tr[u].same = l;
    return;
  }
  int mid = (l + r) / 2;
  init(u * 2, l, mid);
  init(u * 2 + 1, mid + 1, r);
}
void propagate(int u, int l, int r) {
  if (tr[u].lazy > 0) {
    assert(tr[u].same > 0);
    long long n = r - l + 1;
    tr[u].sum += n * tr[u].lazy;
    if (l != r) {
      tr[u * 2].lazy += tr[u].lazy;
      tr[u * 2 + 1].lazy += tr[u].lazy;
      tr[u * 2].same = tr[u].same;
      tr[u * 2 + 1].same = tr[u].same;
    }
    tr[u].lazy = 0;
  }
}
void update(int u, int l, int r, int x, int y, int v) {
  propagate(u, l, r);
  if (y < l || r < x) return;
  if (x <= l && r <= y && tr[u].same > 0) {
    tr[u].lazy += abs(v - tr[u].same);
    tr[u].same = v;
    propagate(u, l, r);
    return;
  }
  int mid = (l + r) / 2;
  update(u * 2, l, mid, x, y, v);
  update(u * 2 + 1, mid + 1, r, x, y, v);
  tr[u].sum = tr[u * 2].sum + tr[u * 2 + 1].sum;
  tr[u].same = 0;
  if (tr[u * 2].same == tr[u * 2 + 1].same) tr[u].same = tr[u * 2].same;
}
long long query(int u, int l, int r, int x, int y) {
  propagate(u, l, r);
  if (y < l || r < x) return 0;
  if (x <= l && r <= y) return tr[u].sum;
  int mid = (l + r) / 2;
  return query(u * 2, l, mid, x, y) + query(u * 2 + 1, mid + 1, r, x, y);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  init(1, 1, n);
  while (m--) {
    int tp, l, r;
    cin >> tp >> l >> r;
    if (tp == 1) {
      int x;
      cin >> x;
      update(1, 1, n, l, r, x);
    } else {
      cout << query(1, 1, n, l, r) << "\n";
    }
  }
  return 0;
}
