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;
int N, Q, op, l, r, v;
long long tree[100100 * 4], xtra[100100 * 4], col[100100 * 4];
int b[100100 * 4], e[100100 * 4];
void init(int n, int beg, int end) {
  b[n] = beg;
  e[n] = end;
  if (beg == end) {
    col[n] = beg;
    return;
  }
  int m = beg + (end - beg) / 2;
  init(2 * n, beg, m);
  init(2 * n + 1, m + 1, end);
}
void push(int n) {
  xtra[2 * n] += xtra[n];
  tree[2 * n] += (e[2 * n] - b[2 * n] + 1) * xtra[n];
  xtra[2 * n + 1] += xtra[n];
  tree[2 * n + 1] += (e[2 * n + 1] - b[2 * n + 1] + 1) * xtra[n];
  xtra[n] = 0;
  if (col[n]) col[2 * n] = col[2 * n + 1] = col[n];
  col[n] = 0;
}
void paint(int n) {
  if (col[n]) {
    xtra[n] += abs(v - col[n]);
    tree[n] += (e[n] - b[n] + 1) * abs(v - col[n]);
    col[n] = v;
    return;
  }
  push(n);
  paint(2 * n);
  paint(2 * n + 1);
  tree[n] = tree[2 * n] + tree[2 * n + 1];
}
void update(int n) {
  if (b[n] > r || e[n] < l) return;
  if (b[n] >= l && e[n] <= r) {
    paint(n);
    col[n] = v;
    return;
  }
  push(n);
  update(2 * n);
  update(2 * n + 1);
  tree[n] = tree[2 * n] + tree[2 * n + 1];
}
long long query(int n) {
  if (b[n] > r || e[n] < l) return 0;
  if (b[n] >= l && e[n] <= r) return tree[n];
  push(n);
  return query(2 * n) + query(2 * n + 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> Q;
  init(1, 1, N);
  while (Q--) {
    cin >> op >> l >> r;
    if (op == 1) {
      cin >> v;
      update(1);
    } else
      cout << query(1) << "\n";
  }
}
