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 int maxn = 1e5 + 20;
int n, m;
long long sum[maxn << 2], add[maxn << 2];
int col[maxn << 2];
void update(int p) {
  sum[p] = sum[((p) << 1)] + sum[((p) << 1 | 1)];
  col[p] = col[((p) << 1)] == col[((p) << 1 | 1)] ? col[((p) << 1)] : -1;
}
void pushdown(int p, int l, int r) {
  int mid = (l + r) >> 1;
  if (add[p]) {
    sum[((p) << 1)] += add[p] * (mid - l + 1), add[((p) << 1)] += add[p];
    sum[((p) << 1 | 1)] += add[p] * (r - mid), add[((p) << 1 | 1)] += add[p];
    add[p] = 0;
  }
  if (col[p] != -1) col[((p) << 1)] = col[((p) << 1 | 1)] = col[p];
}
void change(int l, int r, int c, int ll, int rr, int s) {
  if (l == ll && r == rr) {
    if (col[s] != -1) {
      sum[s] += llabs(c - col[s]) * (r - l + 1);
      add[s] += abs(c - col[s]), col[s] = c;
    } else {
      int mid = (ll + rr) >> 1;
      pushdown(s, ll, rr);
      change(l, mid, c, ll, mid, ((s) << 1));
      change(mid + 1, r, c, mid + 1, rr, ((s) << 1 | 1));
      update(s);
    }
  } else {
    int mid = (ll + rr) >> 1;
    pushdown(s, ll, rr);
    if (r <= mid)
      change(l, r, c, ll, mid, ((s) << 1));
    else if (l > mid)
      change(l, r, c, mid + 1, rr, ((s) << 1 | 1));
    else {
      change(l, mid, c, ll, mid, ((s) << 1));
      change(mid + 1, r, c, mid + 1, rr, ((s) << 1 | 1));
    }
    update(s);
  }
}
long long query(int l, int r, int ll, int rr, int s) {
  if (l == ll && r == rr) return sum[s];
  int mid = (ll + rr) >> 1;
  pushdown(s, ll, rr);
  if (r <= mid)
    return query(l, r, ll, mid, ((s) << 1));
  else if (l > mid)
    return query(l, r, mid + 1, rr, ((s) << 1 | 1));
  else
    return query(l, mid, ll, mid, ((s) << 1)) +
           query(mid + 1, r, mid + 1, rr, ((s) << 1 | 1));
}
void build(int l, int r, int s) {
  if (l == r)
    col[s] = l;
  else {
    int mid = (l + r) >> 1;
    build(l, mid, ((s) << 1));
    build(mid + 1, r, ((s) << 1 | 1));
    update(s);
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin >> n >> m;
  build(1, n, 1);
  int tp, l, r, x;
  while (m--) {
    std::cin >> tp >> l >> r;
    if (tp == 1)
      std::cin >> x, change(l, r, x, 1, n, 1);
    else
      std::cout << query(l, r, 1, n, 1) << std::endl;
  }
  return 0;
}
