#include <bits/stdc++.h>
using namespace std;
const long long N = 5e3 + 5;
vector<pair<long long, pair<long long, long long>>> H;
vector<pair<pair<long long, long long>, long long>> V;
vector<pair<long long, long long>> s[2 * N];
pair<pair<long long, long long>, pair<long long, long long>> a[N];
long long T[2 * N];
long long f[2 * N];
long long g[2 * N];
long long sg[2 * N];
long long n, ans;
void update(long long i, long long val) {
  while (i <= 10001) {
    T[i] += val;
    i += i & -i;
  }
}
long long get(long long i) {
  long long res = 0;
  while (i) {
    res += T[i];
    i -= i & -i;
  }
  return res;
}
void solve() {
  for (long long i = 0, k = 0; i < H.size(); i++) {
    while (k < V.size() && V[k].first.first <= H[i].first) {
      s[i].push_back(make_pair(V[k].second, 1));
      long long l = i;
      long long r = H.size();
      while (r - l > 1) {
        long long mid = (l + r) >> 1;
        if (V[k].first.second >= H[mid].first)
          l = mid;
        else
          r = mid;
      }
      s[r].push_back(make_pair(V[k].second, -1));
      k++;
    }
    for (pair<long long, long long> x : s[i]) {
      update(x.first, x.second);
    }
    for (long long j = i + 1; j <= H.size(); j++) {
      for (pair<long long, long long> x : s[j]) {
        update(x.first, x.second);
      }
      long long l = max(H[i].second.first, H[j].second.first);
      long long r = min(H[i].second.second, H[j].second.second);
      if (l <= r) {
        long long num = get(r) - get(l - 1);
        ans += num * (num - 1) / 2;
      }
    }
    for (long long j = i + 1; j <= H.size(); j++) {
      for (pair<long long, long long> x : s[j]) {
        update(x.first, -x.second);
      }
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i].first.first >> a[i].second.first >> a[i].first.second >>
        a[i].second.second;
    a[i].first.first += 5001;
    a[i].first.second += 5001;
    a[i].second.first += 5001;
    a[i].second.second += 5001;
    if (a[i].first.first > a[i].first.second)
      swap(a[i].first.first, a[i].first.second);
    if (a[i].second.first > a[i].second.second)
      swap(a[i].second.first, a[i].second.second);
    if (a[i].first.first == a[i].first.second) {
      V.push_back(make_pair(a[i].second, a[i].first.first));
    } else
      H.push_back(make_pair(a[i].second.first, a[i].first));
  }
  sort(V.begin(), V.end());
  sort(H.begin(), H.end());
  solve();
  cout << ans;
  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 long long N = 5e3 + 5;
vector<pair<long long, pair<long long, long long>>> H;
vector<pair<pair<long long, long long>, long long>> V;
vector<pair<long long, long long>> s[2 * N];
pair<pair<long long, long long>, pair<long long, long long>> a[N];
long long T[2 * N];
long long f[2 * N];
long long g[2 * N];
long long sg[2 * N];
long long n, ans;
void update(long long i, long long val) {
  while (i <= 10001) {
    T[i] += val;
    i += i & -i;
  }
}
long long get(long long i) {
  long long res = 0;
  while (i) {
    res += T[i];
    i -= i & -i;
  }
  return res;
}
void solve() {
  for (long long i = 0, k = 0; i < H.size(); i++) {
    while (k < V.size() && V[k].first.first <= H[i].first) {
      s[i].push_back(make_pair(V[k].second, 1));
      long long l = i;
      long long r = H.size();
      while (r - l > 1) {
        long long mid = (l + r) >> 1;
        if (V[k].first.second >= H[mid].first)
          l = mid;
        else
          r = mid;
      }
      s[r].push_back(make_pair(V[k].second, -1));
      k++;
    }
    for (pair<long long, long long> x : s[i]) {
      update(x.first, x.second);
    }
    for (long long j = i + 1; j <= H.size(); j++) {
      for (pair<long long, long long> x : s[j]) {
        update(x.first, x.second);
      }
      long long l = max(H[i].second.first, H[j].second.first);
      long long r = min(H[i].second.second, H[j].second.second);
      if (l <= r) {
        long long num = get(r) - get(l - 1);
        ans += num * (num - 1) / 2;
      }
    }
    for (long long j = i + 1; j <= H.size(); j++) {
      for (pair<long long, long long> x : s[j]) {
        update(x.first, -x.second);
      }
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i].first.first >> a[i].second.first >> a[i].first.second >>
        a[i].second.second;
    a[i].first.first += 5001;
    a[i].first.second += 5001;
    a[i].second.first += 5001;
    a[i].second.second += 5001;
    if (a[i].first.first > a[i].first.second)
      swap(a[i].first.first, a[i].first.second);
    if (a[i].second.first > a[i].second.second)
      swap(a[i].second.first, a[i].second.second);
    if (a[i].first.first == a[i].first.second) {
      V.push_back(make_pair(a[i].second, a[i].first.first));
    } else
      H.push_back(make_pair(a[i].second.first, a[i].first));
  }
  sort(V.begin(), V.end());
  sort(H.begin(), H.end());
  solve();
  cout << ans;
  return 0;
}

```