#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const int N = 5e5 + 5;
pair<pair<int, int>, int> a[N][2];
int cnt[2], z, n, i, j;
bitset<2501> b[5001];
long long F[10000];
void calc() {
  for (i = 2; i < 10000; i++) {
    F[i] = F[i - 1] + i - 1;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  calc();
  cin >> n;
  for (int i = 0, x1, x2, y1, y2; i < n; i++) {
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 == x2) {
      a[cnt[0]++][0] = {{min(y1, y2), max(y1, y2)}, x1};
    } else {
      a[cnt[1]++][1] = {{min(x1, x2), max(x1, x2)}, y1};
    }
  }
  if (cnt[0] < cnt[1]) {
    for (i = 0; i < cnt[1]; i++) {
      swap(a[i][0], a[i][1]);
    }
    swap(cnt[0], cnt[1]);
  }
  for (i = 0; i < cnt[0]; i++)
    for (j = 0; j < cnt[1]; j++) {
      if (a[i][0].second >= a[j][1].first.first &&
          a[i][0].second <= a[j][1].first.second &&
          a[j][1].second >= a[i][0].first.first &&
          a[j][1].second <= a[i][0].first.second)
        b[i][j] = 1;
    }
  long long ans = 0;
  for (i = 0; i < cnt[0]; i++)
    for (j = i + 1; j < cnt[0]; j++) {
      z = (b[i] & b[j]).count();
      ans += F[z];
    }
  cout << ans;
}

### Prompt

Your task is to create 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const int N = 5e5 + 5;
pair<pair<int, int>, int> a[N][2];
int cnt[2], z, n, i, j;
bitset<2501> b[5001];
long long F[10000];
void calc() {
  for (i = 2; i < 10000; i++) {
    F[i] = F[i - 1] + i - 1;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  calc();
  cin >> n;
  for (int i = 0, x1, x2, y1, y2; i < n; i++) {
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 == x2) {
      a[cnt[0]++][0] = {{min(y1, y2), max(y1, y2)}, x1};
    } else {
      a[cnt[1]++][1] = {{min(x1, x2), max(x1, x2)}, y1};
    }
  }
  if (cnt[0] < cnt[1]) {
    for (i = 0; i < cnt[1]; i++) {
      swap(a[i][0], a[i][1]);
    }
    swap(cnt[0], cnt[1]);
  }
  for (i = 0; i < cnt[0]; i++)
    for (j = 0; j < cnt[1]; j++) {
      if (a[i][0].second >= a[j][1].first.first &&
          a[i][0].second <= a[j][1].first.second &&
          a[j][1].second >= a[i][0].first.first &&
          a[j][1].second <= a[i][0].first.second)
        b[i][j] = 1;
    }
  long long ans = 0;
  for (i = 0; i < cnt[0]; i++)
    for (j = i + 1; j < cnt[0]; j++) {
      z = (b[i] & b[j]).count();
      ans += F[z];
    }
  cout << ans;
}

```