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>
const double pi = acos(-1);
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const int MAXN = 1e6 + 5;
const double eps = 1e-9;
using namespace std;
struct Node {
  long long lazy, val, col;
  Node() : lazy(0), val(0), col(0){};
};
Node tree[MAXN];
void build(int l, int r, int k) {
  if (l == r) {
    tree[k].lazy = 0;
    tree[k].val = 0;
    tree[k].col = l + 1;
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, k << 1);
  build(m + 1, r, k << 1 | 1);
}
void prop(int k, long long l, long long r) {
  tree[k].val += (r - l + 1) * tree[k].lazy;
  if (l != r) {
    tree[k * 2].lazy += tree[k].lazy;
    tree[k * 2 + 1].lazy += tree[k].lazy;
  }
  tree[k].lazy = 0;
}
long long qry(int l, int r, int k, int a, int b) {
  prop(k, l, r);
  if (r < a || b < l) return 0;
  if (a <= l && r <= b) return tree[k].val;
  int m = (l + r) / 2;
  return qry(l, m, k * 2, a, b) + qry(m + 1, r, k * 2 + 1, a, b);
}
void upd(int l, int r, int k, int a, int b, long long x) {
  prop(k, l, r);
  if (r < a || b < l) return;
  if (a <= l && r <= b && tree[k].col) {
    tree[k].lazy += abs(tree[k].col - x);
    prop(k, l, r);
    tree[k].col = x;
    return;
  }
  if (tree[k].col) tree[k * 2].col = tree[k * 2 + 1].col = tree[k].col;
  int m = (l + r) / 2;
  upd(l, m, k * 2, a, b, x);
  upd(m + 1, r, k * 2 + 1, a, b, x);
  if (tree[k * 2].col == tree[k * 2 + 1].col)
    tree[k].col = tree[k * 2].col;
  else
    tree[k].col = 0;
  tree[k].val = tree[k * 2].val + tree[k * 2 + 1].val;
}
int main() {
  int n, m;
  scanf("%d", &(n)), scanf("%d", &(m));
  build(0, n - 1, 1);
  while (m--) {
    int t, l, r, x;
    scanf("%d", &(t)), scanf("%d", &(l)), scanf("%d", &(r));
    if (t == 1) {
      scanf("%d", &(x));
      upd(0, n - 1, 1, l - 1, r - 1, x);
    } else
      printf("%I64d\n", (qry(0, n - 1, 1, l - 1, r - 1)));
  }
  return 0;
}
