#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 10;
int f[N];
void upd(int pos, int x) {
  for (; pos < N; pos |= pos + 1) f[pos] += x;
}
int get(int pos) {
  int res = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) res += f[pos];
  return res;
}
int get(int l, int r) { return get(r) - get(l - 1); }
const int D = 5005;
void solve() {
  int n;
  cin >> n;
  vector<vector<pair<int, int>>> H(N), V(N);
  for (int i = 0; i < n; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += D, x2 += D, y1 += D, y2 += D;
    if (x1 == x2)
      V[x1].push_back({min(y1, y2), max(y1, y2)});
    else
      H[y1].push_back({min(x1, x2), max(x1, x2)});
  }
  long long ans = 0;
  for (int y = 0; y < N; y++) {
    for (auto p : H[y]) {
      vector<vector<int>> U(N);
      memset(f, 0, sizeof f);
      int l = p.first, r = p.second;
      for (int x = l; x <= r; x++) {
        for (auto q : V[x]) {
          if (q.first <= y && q.second > y) {
            U[q.second].push_back(x);
            upd(x, 1);
          }
        }
      }
      for (int y1 = y + 1; y1 < N; y1++) {
        for (auto q : H[y1]) {
          int count = get(q.first, q.second);
          ans += (count * 1LL * (count - 1)) / 2;
        }
        for (int x : U[y1]) {
          upd(x, -1);
        }
      }
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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;
const int N = 1e4 + 10;
int f[N];
void upd(int pos, int x) {
  for (; pos < N; pos |= pos + 1) f[pos] += x;
}
int get(int pos) {
  int res = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) res += f[pos];
  return res;
}
int get(int l, int r) { return get(r) - get(l - 1); }
const int D = 5005;
void solve() {
  int n;
  cin >> n;
  vector<vector<pair<int, int>>> H(N), V(N);
  for (int i = 0; i < n; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += D, x2 += D, y1 += D, y2 += D;
    if (x1 == x2)
      V[x1].push_back({min(y1, y2), max(y1, y2)});
    else
      H[y1].push_back({min(x1, x2), max(x1, x2)});
  }
  long long ans = 0;
  for (int y = 0; y < N; y++) {
    for (auto p : H[y]) {
      vector<vector<int>> U(N);
      memset(f, 0, sizeof f);
      int l = p.first, r = p.second;
      for (int x = l; x <= r; x++) {
        for (auto q : V[x]) {
          if (q.first <= y && q.second > y) {
            U[q.second].push_back(x);
            upd(x, 1);
          }
        }
      }
      for (int y1 = y + 1; y1 < N; y1++) {
        for (auto q : H[y1]) {
          int count = get(q.first, q.second);
          ans += (count * 1LL * (count - 1)) / 2;
        }
        for (int x : U[y1]) {
          upd(x, -1);
        }
      }
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
}

```