#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
class BIT {
 public:
  vector<long long> tree;
  BIT() { tree.resize(10050, 0LL); }
  void upd(long long id, long long v) {
    for (long long i = id; i <= 10050; i += (i & (-i))) tree[i] += v;
  }
  long long qry(long long id) {
    long long ret = 0;
    for (long long i = id; i > 0; i -= (i & (-i))) ret += tree[i];
    return ret;
  }
  long long ans(long long l, long long r) {
    long long ret = qry(r);
    if (l <= 1) return ret;
    ret -= qry(l - 1);
    return ret;
  }
};
class seg {
 public:
  long long x, y, z, w;
  seg(long long _x, long long _y, long long _z, long long _w) {
    x = _x;
    y = _y;
    z = _z;
    w = _w;
  }
};
long long n;
vector<seg> H, V;
const long long off = 5001;
vector<seg> v[10020];
vector<pair<long long, long long> > events;
void sets(long long j, BIT &hei, vector<pair<long long, long long> > &events) {
  for (auto &k : H) {
    if (k.x <= j && k.z >= j) {
      hei.upd(k.y + off, 1);
      events.push_back(make_pair(k.z + off, k.y + off));
    }
  }
  sort(events.begin(), events.end());
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  set<long long> ss;
  for (long long i = 1; i <= n; i++) {
    long long x, y, z, w;
    cin >> x >> y >> z >> w;
    if (x > z) swap(x, z);
    if (y > w) swap(y, w);
    if (x == z) {
      V.push_back(seg(x, y, z, w));
      v[x + off].push_back(seg(x, y, z, w));
      ss.insert(x + off);
    } else {
      H.push_back(seg(x, y, z, w));
    }
  }
  long long ans = 0;
  vector<long long> vv(ss.begin(), ss.end());
  for (long long i = 0; i < vv.size(); i++) {
    vector<seg> cur = v[vv[i]];
    BIT hei;
    vector<pair<long long, long long> > events;
    sets(vv[i] - off, hei, events);
    long long ept = 0;
    for (long long j = i + 1; j < vv.size(); j++) {
      while (ept < events.size() && events[ept].first < vv[j]) {
        long long hh = events[ept].second;
        hei.upd(hh, -1);
        ept++;
      }
      if (hei.qry(10020) == 0) break;
      for (auto &k : v[vv[j]]) {
        for (auto &g : cur) {
          long long A = g.y;
          long long B = g.w;
          A = max(A, k.y);
          B = min(B, k.w);
          if (A > B) {
            continue;
          }
          long long yo = hei.ans(A + off, B + off);
          ans += (yo * (yo - 1)) / 2;
        }
      }
    }
  }
  cout << ans << endl;
  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>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
class BIT {
 public:
  vector<long long> tree;
  BIT() { tree.resize(10050, 0LL); }
  void upd(long long id, long long v) {
    for (long long i = id; i <= 10050; i += (i & (-i))) tree[i] += v;
  }
  long long qry(long long id) {
    long long ret = 0;
    for (long long i = id; i > 0; i -= (i & (-i))) ret += tree[i];
    return ret;
  }
  long long ans(long long l, long long r) {
    long long ret = qry(r);
    if (l <= 1) return ret;
    ret -= qry(l - 1);
    return ret;
  }
};
class seg {
 public:
  long long x, y, z, w;
  seg(long long _x, long long _y, long long _z, long long _w) {
    x = _x;
    y = _y;
    z = _z;
    w = _w;
  }
};
long long n;
vector<seg> H, V;
const long long off = 5001;
vector<seg> v[10020];
vector<pair<long long, long long> > events;
void sets(long long j, BIT &hei, vector<pair<long long, long long> > &events) {
  for (auto &k : H) {
    if (k.x <= j && k.z >= j) {
      hei.upd(k.y + off, 1);
      events.push_back(make_pair(k.z + off, k.y + off));
    }
  }
  sort(events.begin(), events.end());
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  set<long long> ss;
  for (long long i = 1; i <= n; i++) {
    long long x, y, z, w;
    cin >> x >> y >> z >> w;
    if (x > z) swap(x, z);
    if (y > w) swap(y, w);
    if (x == z) {
      V.push_back(seg(x, y, z, w));
      v[x + off].push_back(seg(x, y, z, w));
      ss.insert(x + off);
    } else {
      H.push_back(seg(x, y, z, w));
    }
  }
  long long ans = 0;
  vector<long long> vv(ss.begin(), ss.end());
  for (long long i = 0; i < vv.size(); i++) {
    vector<seg> cur = v[vv[i]];
    BIT hei;
    vector<pair<long long, long long> > events;
    sets(vv[i] - off, hei, events);
    long long ept = 0;
    for (long long j = i + 1; j < vv.size(); j++) {
      while (ept < events.size() && events[ept].first < vv[j]) {
        long long hh = events[ept].second;
        hei.upd(hh, -1);
        ept++;
      }
      if (hei.qry(10020) == 0) break;
      for (auto &k : v[vv[j]]) {
        for (auto &g : cur) {
          long long A = g.y;
          long long B = g.w;
          A = max(A, k.y);
          B = min(B, k.w);
          if (A > B) {
            continue;
          }
          long long yo = hei.ans(A + off, B + off);
          ans += (yo * (yo - 1)) / 2;
        }
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```