#include <bits/stdc++.h>
using namespace std;
const int maxn = 12007;
const int inf = INT_MAX;
const long long inff = 1e18;
const long long mod = 1e9 + 7;
int n, aa, bb, cc, dd, C = 12000;
int d[maxn];
vector<pair<int, int> > hor[maxn];
vector<int> in[maxn], out[maxn];
inline void Clear() { memset(d, 0, sizeof(d)); }
void modify(int x, int ddd) {
  for (int i = x; i <= C; i += (i & (-i))) d[i] += ddd;
}
int query(int x) {
  if (x == 0) return 0;
  int ret = 0;
  for (int i = x; i > 0; i -= i & (-i)) ret += d[i];
  return ret;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> aa >> bb >> cc >> dd;
    aa += 5001, bb += 5001, cc += 5001, dd += 5001;
    if (bb == dd)
      hor[bb].push_back(make_pair(min(aa, cc), max(aa, cc)));
    else {
      in[min(bb, dd)].push_back(aa);
      out[max(bb, dd)].push_back(aa);
    }
  }
  long long ans = 0ll;
  for (int i = 1; i <= 10001; i++) {
    for (int j : in[i]) modify(j, 1);
    for (int j : out[i]) modify(j, -1);
    vector<int> v;
    for (int k = i + 1; k <= 10001; k++) {
      for (pair<int, int> l : hor[k]) {
        for (pair<int, int> j : hor[i]) {
          int L = max(l.first, j.first), r = min(l.second, j.second);
          if (r <= L) continue;
          int N = query(r) - query(L - 1);
          ans = (ans + N * 1ll * (N - 1) / 2ll);
        }
      }
      for (int l : out[k]) {
        if (query(l) - query(l - 1) == 1) modify(l, -1), v.push_back(l);
      }
    }
    for (int l : v) modify(l, 1);
  }
  cout << ans << endl;
}

### 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;
const int maxn = 12007;
const int inf = INT_MAX;
const long long inff = 1e18;
const long long mod = 1e9 + 7;
int n, aa, bb, cc, dd, C = 12000;
int d[maxn];
vector<pair<int, int> > hor[maxn];
vector<int> in[maxn], out[maxn];
inline void Clear() { memset(d, 0, sizeof(d)); }
void modify(int x, int ddd) {
  for (int i = x; i <= C; i += (i & (-i))) d[i] += ddd;
}
int query(int x) {
  if (x == 0) return 0;
  int ret = 0;
  for (int i = x; i > 0; i -= i & (-i)) ret += d[i];
  return ret;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> aa >> bb >> cc >> dd;
    aa += 5001, bb += 5001, cc += 5001, dd += 5001;
    if (bb == dd)
      hor[bb].push_back(make_pair(min(aa, cc), max(aa, cc)));
    else {
      in[min(bb, dd)].push_back(aa);
      out[max(bb, dd)].push_back(aa);
    }
  }
  long long ans = 0ll;
  for (int i = 1; i <= 10001; i++) {
    for (int j : in[i]) modify(j, 1);
    for (int j : out[i]) modify(j, -1);
    vector<int> v;
    for (int k = i + 1; k <= 10001; k++) {
      for (pair<int, int> l : hor[k]) {
        for (pair<int, int> j : hor[i]) {
          int L = max(l.first, j.first), r = min(l.second, j.second);
          if (r <= L) continue;
          int N = query(r) - query(L - 1);
          ans = (ans + N * 1ll * (N - 1) / 2ll);
        }
      }
      for (int l : out[k]) {
        if (query(l) - query(l - 1) == 1) modify(l, -1), v.push_back(l);
      }
    }
    for (int l : v) modify(l, 1);
  }
  cout << ans << endl;
}

```