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 int sN = 1 << 19;
long long v[sN], d[sN], l[sN], r[sN], s[sN], w[sN];
int n, m;
int t, Left, Right;
long long x;
void down(int a, int c1, int c2) {
  d[c1] += d[a];
  d[c2] += d[a];
  s[c1] += d[a] * w[c1];
  s[c2] += d[a] * w[c2];
  d[a] = 0;
  if (v[a] > 0) v[c1] = v[c2] = v[a];
}
long long get(int a) {
  if (Right < l[a] or r[a] < Left) return 0ll;
  if (Left <= l[a] and r[a] <= Right) return s[a];
  int c1 = a << 1, c2 = c1 + 1;
  down(a, c1, c2);
  return get(c1) + get(c2);
}
void update(int a) {
  if (Right < l[a] or r[a] < Left) return;
  if (Left <= l[a] and r[a] <= Right and v[a] > 0) {
    long long inc = x - v[a];
    if (inc < 0) inc = -inc;
    d[a] += inc;
    s[a] += inc * w[a];
    v[a] = x;
    return;
  }
  int c1 = a << 1, c2 = c1 + 1;
  down(a, c1, c2);
  update(c1);
  update(c2);
  s[a] = s[c1] + s[c2];
  if (v[c1] == v[c2])
    v[a] = v[c1];
  else
    v[a] = 0;
}
int main() {
  scanf("%d %d", &n, &m);
  int nn = 1;
  for (; nn < n; nn <<= 1)
    ;
  n = nn--;
  for (int i = 1; i <= n; i++)
    v[nn + i] = l[nn + i] = r[nn + i] = i, w[nn + i] = 1;
  for (int i = nn, j = i << 1; i > 0; i--, j = i << 1)
    l[i] = l[j], r[i] = r[j + 1], w[i] = w[j] << 1;
  for (; m--;) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d %d", &Left, &Right);
      scanf("%I64d", &x);
      update(1);
    } else {
      scanf("%d %d", &Left, &Right);
      printf("%I64d\n", get(1));
    }
  }
  return 0;
}
