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;
using ll = long long;
const int N = 1 << 17;
ll C[2 * N], V[2 * N], L[2 * N], n, m, t, l, r, x;
void push(int x, int lx, int rx) {
  V[x] += (rx - lx + 1) * L[x];
  if (x < N) {
    L[2 * x] += L[x];
    L[2 * x + 1] += L[x];
    if (C[x]) C[2 * x] = C[2 * x + 1] = C[x];
  }
  L[x] = 0;
}
void maintain(int x) {
  V[x] = V[2 * x] + V[2 * x + 1];
  C[x] = C[2 * x] == C[2 * x + 1] ? C[2 * x] : 0;
}
void update(int x, int lx, int rx, int l, int r, ll c) {
  if (lx > r || rx < l) return;
  if (l <= lx && rx <= r && C[x]) {
    L[x] += abs(c - C[x]);
    C[x] = c;
    return;
  }
  int mid = (lx + rx) / 2;
  push(x, lx, rx);
  update(2 * x, lx, mid, l, r, c);
  update(2 * x + 1, mid + 1, rx, l, r, c);
  push(2 * x, lx, mid);
  push(2 * x + 1, mid + 1, rx);
  maintain(x);
}
ll query(int x, int lx, int rx, int l, int r) {
  if (lx > r || rx < l) return 0;
  push(x, lx, rx);
  if (l <= lx && rx <= r) return V[x];
  int mid = (lx + rx) / 2;
  return query(2 * x, lx, mid, l, r) + query(2 * x + 1, mid + 1, rx, l, r);
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) C[N + i] = i;
  while (m--) {
    cin >> t >> l >> r;
    if (t == 1) {
      cin >> x;
      update(1, 0, N - 1, l, r, x);
    } else {
      cout << query(1, 0, N - 1, l, r) << endl;
    }
  }
}
