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 long long A = 100000000000000LL, N = 228228;
long long i, j, n, m, a, l, r, x, t[3][N * 4];
void talk(long long v, long long l, long long r) {
  if (!t[0][v]) return;
  long long mid = (l + r) / 2;
  t[1][v * 2] += t[2][v] * (mid - l + 1),
      t[1][v * 2 + 1] += t[2][v] * (r - mid);
  t[0][v * 2] = t[0][v * 2 + 1] = t[0][v];
  t[2][v * 2] += t[2][v], t[2][v * 2 + 1] += t[2][v], t[2][v] = 0;
}
void modi(long long v, long long l, long long r, long long _l, long long _r,
          long long x) {
  if (_l > _r) return;
  if (l == _l && r == _r && t[0][v]) {
    t[1][v] += labs(t[0][v] - x) * (r - l + 1), t[2][v] += labs(x - t[0][v]),
        t[0][v] = x;
    return;
  }
  talk(v, l, r);
  long long mid = (l + r) / 2;
  modi(v * 2, l, mid, _l, min(_r, mid), x),
      modi(v * 2 + 1, mid + 1, r, max(_l, mid + 1), _r, x);
  if (t[0][v * 2 + 1] == t[0][v * 2])
    t[0][v] = t[0][v * 2];
  else
    t[0][v] = 0;
  t[1][v] = t[1][v * 2] + t[1][v * 2 + 1];
}
long long get(long long v, long long l, long long r, long long _l,
              long long _r) {
  if (_l > _r) return 0;
  if (l == _l && r == _r) return t[1][v];
  talk(v, l, r);
  long long mid = (l + r) / 2;
  return get(v * 2, l, mid, _l, min(_r, mid)) +
         get(v * 2 + 1, mid + 1, r, max(_l, mid + 1), _r);
}
void build(long long v, long long l, long long r) {
  if (l == r) {
    t[0][v] = l;
    return;
  }
  long long mid = (l + r) / 2;
  build(v * 2, l, mid), build(v * 2 + 1, mid + 1, r);
}
int main() {
  cin >> n >> m;
  build(1, 1, n);
  while (m--) {
    scanf("%d%d%d", &a, &l, &r);
    if (a == 1)
      scanf("%d", &x), modi(1, 1, n, l, r, x);
    else
      cout << get(1, 1, n, l, r) << "\n";
  }
}
