#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
const int MAX_N = 10001;
int f[MAX_N];
void upd(int pos, int x) {
  for (; pos < MAX_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 main() {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<vector<pair<int, int>>> hs(MAX_N);
  vector<vector<pair<int, int>>> vs(MAX_N);
  for (int i = 0; i < N; ++i) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a += 5000;
    b += 5000;
    c += 5000;
    d += 5000;
    if (a > c) swap(a, c);
    if (b > d) swap(b, d);
    if (a == c) {
      vs[a].push_back(make_pair(b, d));
    } else {
      hs[b].push_back(make_pair(a, c));
    }
  }
  long long answer = 0;
  for (int y = 0; y < MAX_N; ++y) {
    for (auto lr : hs[y]) {
      int l = lr.first;
      int r = lr.second;
      for (int i = 0; i < MAX_N; ++i) f[i] = 0;
      vector<vector<pair<int, int>>> querys(2 * MAX_N + 5);
      for (int x = l; x <= r; ++x) {
        for (auto ud : vs[x]) {
          int u = ud.first;
          int d = ud.second;
          if (u <= y && y <= d) {
            querys[2 * d + 1].push_back(make_pair(x, x));
            upd(x, 1);
          }
        }
      }
      for (int y2 = y + 1; y2 < MAX_N; ++y2) {
        for (auto lr2 : hs[y2]) {
          querys[2 * y2].push_back(lr2);
        }
      }
      for (int i = 0; i < querys.size(); ++i) {
        if (i % 2 == 0) {
          for (auto lr : querys[i]) {
            int sum = get(lr.second) - get(lr.first - 1);
            answer += sum * (sum - 1) / 2;
          }
        } else {
          for (auto place : querys[i]) {
            upd(place.first, -1);
          }
        }
      }
    }
  }
  cout << answer << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 MOD = 998244353;
const int MAX_N = 10001;
int f[MAX_N];
void upd(int pos, int x) {
  for (; pos < MAX_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 main() {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<vector<pair<int, int>>> hs(MAX_N);
  vector<vector<pair<int, int>>> vs(MAX_N);
  for (int i = 0; i < N; ++i) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a += 5000;
    b += 5000;
    c += 5000;
    d += 5000;
    if (a > c) swap(a, c);
    if (b > d) swap(b, d);
    if (a == c) {
      vs[a].push_back(make_pair(b, d));
    } else {
      hs[b].push_back(make_pair(a, c));
    }
  }
  long long answer = 0;
  for (int y = 0; y < MAX_N; ++y) {
    for (auto lr : hs[y]) {
      int l = lr.first;
      int r = lr.second;
      for (int i = 0; i < MAX_N; ++i) f[i] = 0;
      vector<vector<pair<int, int>>> querys(2 * MAX_N + 5);
      for (int x = l; x <= r; ++x) {
        for (auto ud : vs[x]) {
          int u = ud.first;
          int d = ud.second;
          if (u <= y && y <= d) {
            querys[2 * d + 1].push_back(make_pair(x, x));
            upd(x, 1);
          }
        }
      }
      for (int y2 = y + 1; y2 < MAX_N; ++y2) {
        for (auto lr2 : hs[y2]) {
          querys[2 * y2].push_back(lr2);
        }
      }
      for (int i = 0; i < querys.size(); ++i) {
        if (i % 2 == 0) {
          for (auto lr : querys[i]) {
            int sum = get(lr.second) - get(lr.first - 1);
            answer += sum * (sum - 1) / 2;
          }
        } else {
          for (auto place : querys[i]) {
            upd(place.first, -1);
          }
        }
      }
    }
  }
  cout << answer << endl;
  return 0;
}

```