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 segtree {
  long long N;
  vector<long long> col, sum, add;
  segtree(long long n) {
    N = 1;
    while (N < n) N <<= 1;
    col.assign(2 * N - 1, -1);
    sum.assign(2 * N - 1, 0);
    add.assign(2 * N - 1, 0);
    for (long long i = 0; i < N; i++) col[N - 1 + i] = i + 1;
  }
  void update(long long a, long long b, long long x) {
    update(a, b, 0, N, 0, x, -1);
  }
  long long update(long long a, long long b, long long l, long long r,
                   long long k, long long x, long long c) {
    if (c >= 0) col[k] = c;
    c = col[k];
    if (b <= l or r <= a or col[k] == x) return 0;
    long long m = (l + r) / 2, s = 0;
    if (a <= l and r <= b) {
      if (col[k] >= 0) {
        add[k] += s = abs(col[k] - x);
        s *= r - l;
      } else
        s = update(a, b, l, m, 2 * k + 1, x, c) +
            update(a, b, m, r, 2 * k + 2, x, c);
      col[k] = x;
    } else {
      s = update(a, b, l, m, 2 * k + 1, x, c) +
          update(a, b, m, r, 2 * k + 2, x, c);
      col[k] = -1;
    }
    sum[k] += s;
    return s;
  }
  long long query(long long a, long long b) { return query(a, b, 0, N, 0); }
  long long query(long long a, long long b, long long l, long long r,
                  long long k) {
    if (b <= l or r <= a) return 0;
    if (a <= l and r <= b) return sum[k];
    long long m = (l + r) / 2, s = min(r, b) - max(l, a);
    return query(a, b, l, m, 2 * k + 1) + query(a, b, m, r, 2 * k + 2) +
           add[k] * s;
  }
};
long long n, m, t, l, r, x;
void solve() {
  segtree st(n);
  while (m--) {
    cin >> t >> l >> r;
    --l;
    if (t == 1) {
      cin >> x;
      st.update(l, r, x);
    } else
      cout << st.query(l, r) << endl;
  }
}
void input() { cin >> n >> m; }
signed main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  input();
  solve();
  return 0;
}
