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 abss(long long x) { return x > 0 ? x : -x; }
struct tree {
  int l, r;
  long long sum, sign, lazy;
  bool flag;
} tr[400000 + 1];
int num;
void build(int i, int l, int r) {
  tr[i].l = l;
  tr[i].r = r;
  if (l == r) {
    tr[i].sign = num++;
    tr[i].flag = true;
    tr[i].sum = 0;
  } else {
    int mid = (l + r) / 2;
    build(i << 1, l, mid);
    build((i << 1) + 1, mid + 1, r);
    tr[i].sum = tr[i << 1].sum + tr[(i << 1) + 1].sum;
    tr[i].flag = false;
    tr[i].sign = 0;
  }
  tr[i].lazy = 0;
}
void pushdown(int i) {
  tr[i << 1].lazy += tr[i].lazy;
  tr[(i << 1) + 1].lazy += tr[i].lazy;
  tr[i << 1].sign = tr[(i << 1) + 1].sign = tr[i].sign;
  tr[i << 1].flag = tr[(i << 1) + 1].flag = true;
  tr[i << 1].sum += (tr[i << 1].r - tr[i << 1].l + 1) * tr[i].lazy;
  tr[(i << 1) + 1].sum +=
      (tr[(i << 1) + 1].r - tr[(i << 1) + 1].l + 1) * tr[i].lazy;
  tr[i].lazy = 0;
}
void update(int i, int l, int r, long long v) {
  if (tr[i].flag && tr[i].l == l && tr[i].r == r) {
    tr[i].sum += (tr[i].r - tr[i].l + 1) * abss(tr[i].sign - v);
    tr[i].lazy += abss(tr[i].sign - v);
    tr[i].sign = v;
  } else {
    if (tr[i].lazy) pushdown(i);
    int mid = (tr[i].l + tr[i].r) / 2;
    if (r <= mid)
      update(i << 1, l, r, v);
    else if (l > mid)
      update((i << 1) + 1, l, r, v);
    else {
      update(i << 1, l, mid, v);
      update((i << 1) + 1, mid + 1, r, v);
    }
    tr[i].sum = tr[i << 1].sum + tr[(i << 1) + 1].sum;
    if (tr[i].l == l && tr[i].r == r) {
      tr[i].flag = true;
      tr[i].sign = v;
    } else if (tr[i].sign != v)
      tr[i].flag = false;
  }
}
long long query(int i, int l, int r) {
  if (tr[i].l == l && tr[i].r == r)
    return tr[i].sum;
  else {
    if (tr[i].lazy) pushdown(i);
    int mid = (tr[i].l + tr[i].r) / 2;
    if (r <= mid)
      return query(i << 1, l, r);
    else if (l > mid)
      return query((i << 1) + 1, l, r);
    else
      return query(i << 1, l, mid) + query((i << 1) + 1, mid + 1, r);
  }
}
int main() {
  int n, m, flag, l, r, i;
  long long v;
  scanf("%d %d", &n, &m);
  num = 1;
  build(1, 1, n);
  for (i = 0; i < m; i++) {
    scanf("%d", &flag);
    if (flag == 1) {
      scanf("%d %d %I64d", &l, &r, &v);
      update(1, l, r, v);
    } else {
      scanf("%d %d", &l, &r);
      printf("%I64d\n", query(1, l, r));
    }
  }
  return 0;
}
