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;
struct node {
  long long sum;
  long long val;
  long long color;
};
node T[400040];
int n, m;
void built(int node, int l, int r) {
  if (l == r) {
    T[node].color = l;
    T[node].sum = 0;
    T[node].val = 0;
    return;
  }
  int mid = (l + r) >> 1;
  built(node << 1, l, mid);
  built(node << 1 | 1, mid + 1, r);
}
void update(int node, int l, int r, int u, int v, int col) {
  if ((r < u) || (l > v)) return;
  if ((u <= l) && (r <= v) && (T[node].color)) {
    T[node].sum += abs(col - T[node].color) * (r - l + 1);
    T[node].val += abs(col - T[node].color);
    T[node].color = col;
    return;
  }
  int mid = (l + r) >> 1;
  if (T[node].val > 0) {
    T[node << 1].val += T[node].val;
    T[node << 1].sum += T[node].val * (mid - l + 1);
    T[node << 1 | 1].val += T[node].val;
    T[node << 1 | 1].sum += T[node].val * (r - mid);
    T[node].val = 0;
    T[node << 1].color = T[node].color;
    T[node << 1 | 1].color = T[node].color;
  }
  T[node].color = 0;
  update(node << 1, l, mid, u, v, col);
  update(node << 1 | 1, mid + 1, r, u, v, col);
  T[node].sum = T[node << 1].sum + T[node << 1 | 1].sum;
  if (T[node << 1].color == T[node << 1 | 1].color)
    T[node].color = T[node << 1].color;
}
long long query(int node, int l, int r, int u, int v) {
  long long res = 0;
  if ((v < l) || (r < u)) return 0;
  if ((u <= l) && (r <= v)) {
    return T[node].sum;
  }
  int mid = (l + r) >> 1;
  if (T[node].val > 0) {
    T[node << 1].val += T[node].val;
    T[node << 1].sum += T[node].val * (mid - l + 1);
    T[node << 1 | 1].val += T[node].val;
    T[node << 1 | 1].sum += T[node].val * (r - mid);
    T[node].val = 0;
    T[node << 1].color = T[node].color;
    T[node << 1 | 1].color = T[node].color;
  }
  res += query(node << 1, l, mid, u, v);
  res += query(node << 1 | 1, mid + 1, r, u, v);
  T[node].sum = T[node << 1].sum + T[node << 1 | 1].sum;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  built(1, 1, n);
  int u, v, col, x;
  for (int i = 1; i <= m; i++) {
    cin >> x;
    if (x == 1) {
      cin >> u >> v >> col;
      update(1, 1, n, u, v, col);
    } else if (x == 2) {
      cin >> u >> v;
      cout << query(1, 1, n, u, v) << "\n";
    }
  }
  return 0;
}
