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;
struct nod {
  long long sum, lazy;
  int same;
} tree[100005 * 5];
int n, m, l, r, t, v;
void build_tree(int id, int l, int r) {
  tree[id] = (nod){0, 0, 0};
  if (l == r) {
    tree[id].same = l;
    return;
  }
  build_tree((id << 1), l, ((l + r) >> 1));
  build_tree(((id << 1) | 1), ((l + r) >> 1) + 1, r);
  return;
}
void push_down(int id, int l, int r) {
  if (!tree[id].lazy) return;
  tree[(id << 1)].sum += 1ll * (((l + r) >> 1) - l + 1) * tree[id].lazy;
  tree[(id << 1)].same = tree[id].same;
  tree[(id << 1)].lazy += tree[id].lazy;
  tree[((id << 1) | 1)].sum += 1ll * (r - ((l + r) >> 1)) * tree[id].lazy;
  tree[((id << 1) | 1)].same = tree[id].same;
  tree[((id << 1) | 1)].lazy += tree[id].lazy;
  tree[id].lazy = 0;
  return;
}
void push_up(int id) {
  tree[id].sum = tree[(id << 1)].sum + tree[((id << 1) | 1)].sum;
  if (tree[(id << 1)].same == tree[((id << 1) | 1)].same)
    tree[id].same = tree[(id << 1)].same;
  else
    tree[id].same = 0;
  return;
}
void ins(int id, int l, int r, int ql, int qr, int tt) {
  if (ql <= l && r <= qr && tree[id].same) {
    tree[id].sum += 1ll * (r - l + 1) * abs(tree[id].same - tt);
    tree[id].lazy += 1ll * abs(tree[id].same - tt);
    tree[id].same = tt;
    return;
  }
  if (tree[id].lazy) push_down(id, l, r);
  if (ql <= ((l + r) >> 1)) ins((id << 1), l, ((l + r) >> 1), ql, qr, tt);
  if (((l + r) >> 1) + 1 <= qr)
    ins(((id << 1) | 1), ((l + r) >> 1) + 1, r, ql, qr, tt);
  push_up(id);
  return;
}
long long query(int id, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) {
    return tree[id].sum;
  }
  if (tree[id].lazy) push_down(id, l, r);
  long long sum = 0;
  if (ql <= ((l + r) >> 1)) sum += query((id << 1), l, ((l + r) >> 1), ql, qr);
  if (((l + r) >> 1) + 1 <= qr)
    sum += query(((id << 1) | 1), ((l + r) >> 1) + 1, r, ql, qr);
  return sum;
}
int main() {
  scanf("%d%d", &n, &m);
  build_tree(1, 1, n);
  while (m--) {
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1) {
      scanf("%d", &v);
      ins(1, 1, n, l, r, v);
    } else
      printf("%lld\n", query(1, 1, n, l, r));
  }
}
