#include <bits/stdc++.h>
using namespace std;
struct bit {
  vector<long long> v;
  int n;
  bit(int n) {
    this->n = n;
    v.assign(n, 0);
  }
  int sum(int r) {
    int ret = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) {
      ret += v[r];
    }
    return ret;
  }
  int sum(int l, int r) { return sum(r) - sum(l - 1); }
  void add(int idx, int delta) {
    for (; idx < n; idx = idx | (idx + 1)) v[idx] += delta;
  }
};
int main() {
  std::ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  vector<pair<long long, pair<long long, long long> > > horizontal, vertical;
  bit bb(10001);
  for (int i = 0; i < n; i++) {
    long long a, b, c, d;
    cin >> a >> b >> c >> d;
    a += 5000;
    b += 5000;
    c += 5000;
    d += 5000;
    if (a == c) {
      vertical.emplace_back(a,
                            pair<long long, long long>(min(b, d), max(b, d)));
    } else {
      horizontal.emplace_back(b,
                              pair<long long, long long>(min(a, c), max(a, c)));
    }
  }
  sort(vertical.begin(), vertical.end());
  long long ans = 0;
  for (long long i = 0; i < vertical.size(); i++) {
    long long iX = vertical[i].first;
    long long iyBegin = vertical[i].second.first;
    long long iyEnd = vertical[i].second.second;
    vector<pair<long long, long long> > intercept;
    for (auto h : horizontal) {
      if (h.first < iyBegin || h.first > iyEnd) continue;
      if (iX < h.second.first || iX > h.second.second) continue;
      intercept.emplace_back(h.second.second, h.first);
      bb.add(h.first, 1);
    }
    sort(intercept.begin(), intercept.end());
    long long interceptIdx = 0;
    for (long long j = i + 1; j < vertical.size(); j++) {
      long long jX = vertical[j].first;
      long long jyBegin = vertical[j].second.first;
      long long jyEnd = vertical[j].second.second;
      if (jyBegin > iyEnd || jyEnd < iyBegin) continue;
      while (interceptIdx < intercept.size() &&
             intercept[interceptIdx].first < jX) {
        bb.add(intercept[interceptIdx].second, -1);
        interceptIdx++;
      }
      long long count = bb.sum(jyBegin, jyEnd);
      ans += ((count) * (count - 1)) / 2;
    }
    while (interceptIdx < intercept.size()) {
      bb.add(intercept[interceptIdx].second, -1);
      interceptIdx++;
    }
  }
  cout << ans << "\n";
}

### 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>
using namespace std;
struct bit {
  vector<long long> v;
  int n;
  bit(int n) {
    this->n = n;
    v.assign(n, 0);
  }
  int sum(int r) {
    int ret = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) {
      ret += v[r];
    }
    return ret;
  }
  int sum(int l, int r) { return sum(r) - sum(l - 1); }
  void add(int idx, int delta) {
    for (; idx < n; idx = idx | (idx + 1)) v[idx] += delta;
  }
};
int main() {
  std::ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  vector<pair<long long, pair<long long, long long> > > horizontal, vertical;
  bit bb(10001);
  for (int i = 0; i < n; i++) {
    long long a, b, c, d;
    cin >> a >> b >> c >> d;
    a += 5000;
    b += 5000;
    c += 5000;
    d += 5000;
    if (a == c) {
      vertical.emplace_back(a,
                            pair<long long, long long>(min(b, d), max(b, d)));
    } else {
      horizontal.emplace_back(b,
                              pair<long long, long long>(min(a, c), max(a, c)));
    }
  }
  sort(vertical.begin(), vertical.end());
  long long ans = 0;
  for (long long i = 0; i < vertical.size(); i++) {
    long long iX = vertical[i].first;
    long long iyBegin = vertical[i].second.first;
    long long iyEnd = vertical[i].second.second;
    vector<pair<long long, long long> > intercept;
    for (auto h : horizontal) {
      if (h.first < iyBegin || h.first > iyEnd) continue;
      if (iX < h.second.first || iX > h.second.second) continue;
      intercept.emplace_back(h.second.second, h.first);
      bb.add(h.first, 1);
    }
    sort(intercept.begin(), intercept.end());
    long long interceptIdx = 0;
    for (long long j = i + 1; j < vertical.size(); j++) {
      long long jX = vertical[j].first;
      long long jyBegin = vertical[j].second.first;
      long long jyEnd = vertical[j].second.second;
      if (jyBegin > iyEnd || jyEnd < iyBegin) continue;
      while (interceptIdx < intercept.size() &&
             intercept[interceptIdx].first < jX) {
        bb.add(intercept[interceptIdx].second, -1);
        interceptIdx++;
      }
      long long count = bb.sum(jyBegin, jyEnd);
      ans += ((count) * (count - 1)) / 2;
    }
    while (interceptIdx < intercept.size()) {
      bb.add(intercept[interceptIdx].second, -1);
      interceptIdx++;
    }
  }
  cout << ans << "\n";
}

```