#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f, N = 1e4 + 5;
inline int read() {
  int sum = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum * f;
}
int n, c[N];
vector<pair<int, int> > h[N], v[N];
vector<int> now[N];
void add(int x, int v) {
  for (int i = x; i < N; i += (i & -i)) c[i] += v;
}
int sum(int x) {
  int ans = 0;
  for (int i = x; i > 0; i -= (i & -i)) ans += c[i];
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a += 5001, b += 5001, c += 5001, d += 5001;
    if (a > c) swap(a, c);
    if (b > d) swap(b, d);
    if (a == c)
      v[a].emplace_back(b, d);
    else
      h[b].emplace_back(a, c);
  }
  long long ans = 0;
  for (int y = 1; y < N; y++) {
    for (auto i : h[y]) {
      for (int i = 0; i < N; i++) {
        c[i] = 0;
        now[i].clear();
      }
      for (int x = i.first; x <= i.second; x++) {
        for (auto j : v[x]) {
          if (j.first <= y && j.second > y) {
            add(x, 1);
            now[j.second].emplace_back(x);
          }
        }
      }
      for (int y2 = y + 1; y2 < N; y2++) {
        for (auto j : h[y2]) {
          int tmp = sum(j.second) - sum(j.first - 1);
          ans += 1ll * tmp * (tmp - 1) / 2;
        }
        for (auto j : now[y2]) {
          add(j, -1);
        }
      }
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Please create a solution in Cpp 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 INF = 0x3f3f3f3f, N = 1e4 + 5;
inline int read() {
  int sum = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum * f;
}
int n, c[N];
vector<pair<int, int> > h[N], v[N];
vector<int> now[N];
void add(int x, int v) {
  for (int i = x; i < N; i += (i & -i)) c[i] += v;
}
int sum(int x) {
  int ans = 0;
  for (int i = x; i > 0; i -= (i & -i)) ans += c[i];
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a += 5001, b += 5001, c += 5001, d += 5001;
    if (a > c) swap(a, c);
    if (b > d) swap(b, d);
    if (a == c)
      v[a].emplace_back(b, d);
    else
      h[b].emplace_back(a, c);
  }
  long long ans = 0;
  for (int y = 1; y < N; y++) {
    for (auto i : h[y]) {
      for (int i = 0; i < N; i++) {
        c[i] = 0;
        now[i].clear();
      }
      for (int x = i.first; x <= i.second; x++) {
        for (auto j : v[x]) {
          if (j.first <= y && j.second > y) {
            add(x, 1);
            now[j.second].emplace_back(x);
          }
        }
      }
      for (int y2 = y + 1; y2 < N; y2++) {
        for (auto j : h[y2]) {
          int tmp = sum(j.second) - sum(j.first - 1);
          ans += 1ll * tmp * (tmp - 1) / 2;
        }
        for (auto j : now[y2]) {
          add(j, -1);
        }
      }
    }
  }
  cout << ans << '\n';
  return 0;
}

```