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;
using ll = long long;
using pii = pair<int, int>;
int const N = 100 * 1000 + 16;
int const M = 1000 * 1000 * 1000 + 7;
struct fenwick {
  int n;
  vector<ll> a, b;
  fenwick(int s) {
    n = s;
    a.assign(n + 1, 0);
    b.assign(n + 1, 0);
  }
  void update(vector<ll>& bit, int idx, ll v) {
    for (int i = idx; i <= n; i += (i & -i)) bit[i] += v;
  }
  void update(int l, int r, ll v) {
    update(a, l, v);
    update(a, r + 1, -v);
    update(b, l, -v * (l - 1));
    update(b, r + 1, v * r);
  }
  ll query(vector<ll> const& bit, ll idx) {
    ll sum = 0;
    for (int i = idx; i > 0; i -= (i & -i)) sum += bit[i];
    return sum;
  }
  ll query(int idx) { return query(a, idx) * idx + query(b, idx); }
  ll query(int l, int r) { return query(r) - query(l - 1); }
};
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  fenwick ft(n);
  map<int, pair<int, int>> s;
  for (int i = 1; i <= n; ++i) s[i] = {i, i};
  while (m--) {
    int t, l, r;
    cin >> t >> l >> r;
    if (t & 1) {
      int v;
      cin >> v;
      for (auto it = s.lower_bound(l); it != s.end();) {
        int x, y = it->first, z;
        tie(x, z) = it->second;
        int L = max(l, x);
        int R = min(r, y);
        if (L > R) break;
        s.erase(it++);
        if (x < L) s[L - 1] = {x, z};
        if (R < y) s[y] = {R + 1, z};
        ft.update(L, R, abs(v - z));
      }
      s[r] = {l, v};
    } else {
      cout << ft.query(l, r) << "\n";
    }
  }
}
