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;
int n, m;
long long colour[200005 * 4], sum[200005 * 4], lazy[200005 * 4];
void init(int s, int e, int p) {
  if (s == e)
    colour[p] = s + 1;
  else {
    init(s, (s + e) / 2, p * 2);
    init((s + e) / 2 + 1, e, p * 2 + 1);
  }
}
void relax(int p, int s, int e) {
  if (!lazy[p]) return;
  colour[p * 2] = colour[p * 2 + 1] = colour[p];
  int mid = (s + e) / 2;
  sum[p * 2] += lazy[p] * (mid - s + 1) * 1LL;
  sum[p * 2 + 1] += lazy[p] * (e - mid) * 1LL;
  if (s != e) {
    lazy[p * 2] += lazy[p];
    lazy[p * 2 + 1] += lazy[p];
  }
  lazy[p] = 0;
}
void update(int p, int s, int e, int from, int to, long long x) {
  relax(p, s, e);
  if (s > to || from > e) return;
  if (from <= s && e <= to) {
    if (colour[p]) {
      sum[p] += abs(colour[p] - x) * (e - s + 1) * 1LL;
      lazy[p] += abs(colour[p] - x);
      colour[p] = x;
    } else {
      int mid = (s + e) / 2;
      update(p * 2, s, mid, from, to, x);
      update(p * 2 + 1, mid + 1, e, from, to, x);
      sum[p] = sum[p * 2] + sum[p * 2 + 1];
      colour[p] = x;
    }
    return;
  }
  int mid = (s + e) / 2;
  update(p * 2, s, mid, from, to, x);
  update(p * 2 + 1, mid + 1, e, from, to, x);
  sum[p] = sum[p * 2] + sum[p * 2 + 1];
  colour[p] = colour[p * 2] == colour[p * 2 + 1] ? colour[p * 2] : 0;
}
long long query(int p, int s, int e, int from, int to) {
  relax(p, s, e);
  if (s > to || from > e) return 0;
  if (from <= s && e <= to) return sum[p];
  int mid = (s + e) / 2;
  return query(p * 2, s, mid, from, to) +
         query(p * 2 + 1, mid + 1, e, from, to);
}
int main() {
  scanf("%d%d", &n, &m);
  init(0, n - 1, 1);
  int t, from, to;
  long long x;
  while (m--) {
    scanf("%d%d%d", &t, &from, &to);
    if (t == 2)
      printf("%I64d\n", query(1, 0, n - 1, from - 1, to - 1));
    else {
      cin >> x;
      update(1, 0, n - 1, from - 1, to - 1, x);
    }
  }
  return 0;
}
