#include <bits/stdc++.h>
using namespace std;
;
const long long N = 5e3 + 10;
vector<bitset<N> > v;
vector<pair<long long, pair<long long, long long> > > hor, ver;
long long n, cnt, y11, y2, x1, x2;
long long ans;
bool check(long long x, pair<long long, long long> a) {
  return a.first <= x && x <= a.second;
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> x1 >> y11 >> x2 >> y2;
    if (x1 > x2) swap(x1, x2);
    if (y11 > y2) swap(y11, y2);
    if (x1 == x2)
      ver.push_back({x1, {y11, y2}});
    else
      hor.push_back({y11, {x1, x2}});
  }
  if ((long long)((hor).size()) > (long long)((ver).size())) swap(hor, ver);
  v.resize((long long)((hor).size()));
  for (long long i = 0; i < (long long)((hor).size()); i++) {
    for (long long j = 0; j < (long long)((ver).size()); j++) {
      v[i][j] = check(hor[i].first, ver[j].second) &&
                check(ver[j].first, hor[i].second);
    }
  }
  for (long long i = 0; i < (long long)((v).size()); i++) {
    for (long long j = i + 1; j < (long long)((v).size()); j++) {
      cnt = (v[i] & v[j]).count();
      ans += cnt * (cnt - 1) / 2;
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  clock_t start, end;
  start = clock();
  cout << fixed << setprecision(15);
  while (t--) {
    solve();
  }
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
}

### Prompt

Create a solution in CPP for 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 + 10;
vector<bitset<N> > v;
vector<pair<long long, pair<long long, long long> > > hor, ver;
long long n, cnt, y11, y2, x1, x2;
long long ans;
bool check(long long x, pair<long long, long long> a) {
  return a.first <= x && x <= a.second;
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> x1 >> y11 >> x2 >> y2;
    if (x1 > x2) swap(x1, x2);
    if (y11 > y2) swap(y11, y2);
    if (x1 == x2)
      ver.push_back({x1, {y11, y2}});
    else
      hor.push_back({y11, {x1, x2}});
  }
  if ((long long)((hor).size()) > (long long)((ver).size())) swap(hor, ver);
  v.resize((long long)((hor).size()));
  for (long long i = 0; i < (long long)((hor).size()); i++) {
    for (long long j = 0; j < (long long)((ver).size()); j++) {
      v[i][j] = check(hor[i].first, ver[j].second) &&
                check(ver[j].first, hor[i].second);
    }
  }
  for (long long i = 0; i < (long long)((v).size()); i++) {
    for (long long j = i + 1; j < (long long)((v).size()); j++) {
      cnt = (v[i] & v[j]).count();
      ans += cnt * (cnt - 1) / 2;
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  clock_t start, end;
  start = clock();
  cout << fixed << setprecision(15);
  while (t--) {
    solve();
  }
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
}

```