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 sum[263000], pro[263000], a[263000];
long long S, E, V;
void ini(long long s, long long e, long long node) {
  if (s == e) {
    a[node] = s;
    return;
  }
  long long lft = node << 1, rht = (node << 1) | 1, mid = (s + e) >> 1;
  ini(s, mid, lft);
  ini(mid + 1, e, rht);
  return;
}
void update(long long s, long long e, long long node) {
  if (a[node]) {
    pro[node] += abs(V - a[node]);
    sum[node] += abs(V - a[node]) * (e - s + 1);
    a[node] = 0;
    return;
  }
  long long lft = node << 1, rht = (node << 1) | 1, mid = (s + e) >> 1;
  update(s, mid, lft);
  update(mid + 1, e, rht);
  sum[node] = sum[lft] + sum[rht] + pro[node] * (e - s + 1);
  return;
}
void update(long long s, long long e, long long node, long long carry) {
  if (carry) a[node] = carry;
  if (S > e || s > E) return;
  if (S <= s && e <= E) {
    update(s, e, node);
    a[node] = V;
    return;
  }
  long long lft = node << 1, rht = (node << 1) | 1, mid = (s + e) >> 1;
  if (a[node]) carry = a[node];
  update(s, mid, lft, carry);
  update(mid + 1, e, rht, carry);
  a[node] = 0;
  sum[node] = sum[lft] + sum[rht] + pro[node] * (e - s + 1);
  return;
}
long long ans = 0;
void qry(long long s, long long e, long long node, long long carry) {
  if (S > e || s > E) return;
  if (S <= s && e <= E) {
    ans += sum[node];
    ans += (e - s + 1) * carry;
    return;
  }
  long long lft = node << 1, rht = (node << 1) | 1, mid = (s + e) >> 1;
  carry += pro[node];
  qry(s, mid, lft, carry);
  qry(mid + 1, e, rht, carry);
  return;
}
int main() {
  long long n, m;
  scanf("%I64d %I64d", &n, &m);
  memset(sum, 0, sizeof(sum));
  memset(pro, 0, sizeof(pro));
  memset(a, 0, sizeof(a));
  ini(1, n, 1);
  long long type;
  while (m--) {
    scanf("%I64d %I64d %I64d", &type, &S, &E);
    if (type == 1) {
      scanf("%I64d", &V);
      update(1, n, 1, 0);
    } else {
      ans = 0;
      qry(1, n, 1, 0);
      printf("%I64d", ans);
      printf("\n");
    }
  }
  return 0;
}
