#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 5001, P = 2 * N + 1;
struct binary_indexed_tree : vector<int> {
  int n;
  binary_indexed_tree(int N) : n(N) { resize(N, 0); }
  void update(int pos) {
    for (++pos; pos < n; pos += pos & -pos) ++at(pos);
  }
  int get(int pos) {
    int ans = 0;
    for (++pos; pos > 0; pos -= pos & -pos) ans += at(pos);
    return ans;
  }
};
int m, n, p;
vector<tuple<int, int, int>> h, v;
ll ans;
inline ll C2(ll k) { return k *= k - 1, k >>= 1; }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr),
      cin >> n;
  for (int _ = int(0); _ < int(n); ++_) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2)
      v.emplace_back(x1 + N, y1 + N, y2 + N);
    else
      h.emplace_back(x1 + N, x2 + N, y1 + N);
  }
  sort(v.begin(), v.end()), sort(h.begin(), h.end()), m = v.size(), p = n - m;
  for (int i = int(0); i < int(m); ++i) {
    auto [x, y1, y2] = v[i];
    binary_indexed_tree t(P);
    int j = 0;
    for (int k = int(0); k < int(i); ++k) {
      auto [x3, y3, y4] = v[k];
      for (; j < p; ++j) {
        auto [x1, x2, y] = h[j];
        if (x1 > x3) break;
        if (x <= x2 and y >= y1 and y <= y2) t.update(y);
      }
      ans += C2(t.get(y4) - t.get(y3 - 1));
    }
  }
  cout << ans;
}

### Prompt

Construct a CPP code solution to the problem outlined:

There are n segments drawn on a plane; the i-th segment connects two points (x_{i, 1}, y_{i, 1}) and (x_{i, 2}, y_{i, 2}). Each segment is non-degenerate, and is either horizontal or vertical — formally, for every i ∈ [1, n] either x_{i, 1} = x_{i, 2} or y_{i, 1} = y_{i, 2} (but only one of these conditions holds). Only segments of different types may intersect: no pair of horizontal segments shares any common points, and no pair of vertical segments shares any common points.

We say that four segments having indices h_1, h_2, v_1 and v_2 such that h_1 < h_2 and v_1 < v_2 form a rectangle if the following conditions hold:

  * segments h_1 and h_2 are horizontal; 
  * segments v_1 and v_2 are vertical; 
  * segment h_1 intersects with segment v_1; 
  * segment h_2 intersects with segment v_1; 
  * segment h_1 intersects with segment v_2; 
  * segment h_2 intersects with segment v_2. 



Please calculate the number of ways to choose four segments so they form a rectangle. Note that the conditions h_1 < h_2 and v_1 < v_2 should hold.

Input

The first line contains one integer n (1 ≤ n ≤ 5000) — the number of segments.

Then n lines follow. The i-th line contains four integers x_{i, 1}, y_{i, 1}, x_{i, 2} and y_{i, 2} denoting the endpoints of the i-th segment. All coordinates of the endpoints are in the range [-5000, 5000].

It is guaranteed that each segment is non-degenerate and is either horizontal or vertical. Furthermore, if two segments share a common point, one of these segments is horizontal, and another one is vertical.

Output

Print one integer — the number of ways to choose four segments so they form a rectangle.

Examples

Input


7
-1 4 -1 -2
6 -1 -2 -1
-2 3 6 3
2 -2 2 4
4 -1 4 3
5 3 5 1
5 2 1 2


Output


7


Input


5
1 5 1 0
0 1 5 1
5 4 0 4
4 2 4 0
4 3 4 5


Output


0

Note

The following pictures represent sample cases:

<image> <image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 5001, P = 2 * N + 1;
struct binary_indexed_tree : vector<int> {
  int n;
  binary_indexed_tree(int N) : n(N) { resize(N, 0); }
  void update(int pos) {
    for (++pos; pos < n; pos += pos & -pos) ++at(pos);
  }
  int get(int pos) {
    int ans = 0;
    for (++pos; pos > 0; pos -= pos & -pos) ans += at(pos);
    return ans;
  }
};
int m, n, p;
vector<tuple<int, int, int>> h, v;
ll ans;
inline ll C2(ll k) { return k *= k - 1, k >>= 1; }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr),
      cin >> n;
  for (int _ = int(0); _ < int(n); ++_) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2)
      v.emplace_back(x1 + N, y1 + N, y2 + N);
    else
      h.emplace_back(x1 + N, x2 + N, y1 + N);
  }
  sort(v.begin(), v.end()), sort(h.begin(), h.end()), m = v.size(), p = n - m;
  for (int i = int(0); i < int(m); ++i) {
    auto [x, y1, y2] = v[i];
    binary_indexed_tree t(P);
    int j = 0;
    for (int k = int(0); k < int(i); ++k) {
      auto [x3, y3, y4] = v[k];
      for (; j < p; ++j) {
        auto [x1, x2, y] = h[j];
        if (x1 > x3) break;
        if (x <= x2 and y >= y1 and y <= y2) t.update(y);
      }
      ans += C2(t.get(y4) - t.get(y3 - 1));
    }
  }
  cout << ans;
}

```