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;
long long int n, m, kd[4 * ((int)1e5 + 5)], color[4 * ((int)1e5 + 5)],
    lazy[4 * ((int)1e5 + 5)], c, x, y, v;
void init(long long int root, long long int bas, long long int son) {
  if (bas == son) {
    kd[root] = bas;
    return;
  }
  init((root << 1), bas, ((bas + son) >> 1));
  init(((root << 1) | 1), ((bas + son) >> 1) + 1, son);
}
long long int query(long long int root, long long int bas, long long int son,
                    long long int x, long long int y) {
  if (son < x or y < bas) return 0;
  if (x <= bas and son <= y) return color[root] + lazy[root] * (son - bas + 1);
  if (kd[root]) {
    lazy[(root << 1)] += lazy[root];
    lazy[((root << 1) | 1)] += lazy[root];
    kd[(root << 1)] = kd[root];
    kd[((root << 1) | 1)] = kd[root];
    lazy[root] = kd[root] = 0;
  }
  long long int ret =
      query((root << 1), bas, ((bas + son) >> 1), x, y) +
      query(((root << 1) | 1), ((bas + son) >> 1) + 1, son, x, y);
  color[root] = color[(root << 1)] + color[((root << 1) | 1)] +
                lazy[(root << 1)] * (((bas + son) >> 1) - bas + 1) +
                lazy[((root << 1) | 1)] * (son - (((bas + son) >> 1) + 1) + 1);
  if (kd[(root << 1)] == kd[((root << 1) | 1)]) kd[root] = kd[(root << 1)];
  return ret;
}
long long int update(long long int root, long long int bas, long long int son,
                     long long int x, long long int y, long long int v) {
  if (son < x or y < bas) return color[root] + lazy[root] * (son - bas + 1);
  if (x <= bas and son <= y and kd[root]) {
    lazy[root] += abs(kd[root] - v);
    kd[root] = v;
    return color[root] + lazy[root] * (son - bas + 1);
  }
  if (kd[root]) {
    lazy[(root << 1)] += lazy[root];
    lazy[((root << 1) | 1)] += lazy[root];
    kd[(root << 1)] = kd[root];
    kd[((root << 1) | 1)] = kd[root];
    lazy[root] = kd[root] = 0;
  }
  color[root] = update((root << 1), bas, ((bas + son) >> 1), x, y, v) +
                update(((root << 1) | 1), ((bas + son) >> 1) + 1, son, x, y, v);
  if (kd[(root << 1)] == kd[((root << 1) | 1)]) kd[root] = kd[(root << 1)];
  return color[root];
}
int main() {
  cin >> n >> m;
  init(1, 1, n);
  for (int i = 1; i <= m; i++) {
    cin >> c >> x >> y;
    if (c == 1) {
      cin >> v;
      update(1, 1, n, x, y, v);
    } else
      cout << query(1, 1, n, x, y) << '\n';
  }
  return 0;
}
