#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int grand(int x) { return uniform_int_distribution<int>(0, x - 1)(rng); }
long long gcd(long long x, long long y) {
  if (x < y) return gcd(y, x);
  if (y == 0) return x;
  return gcd(y, x % y);
}
const long long mod = 1e9 + 7;
long long modexp(long long x, long long ex) {
  long long ans = 1ll;
  while (ex > 0) {
    if (ex & 1ll) ans = (ans * x) % mod;
    ex >>= 1ll;
    x = (x * x) % mod;
  }
  return ans;
}
const int maxn = 1e4 + 7;
const long long inf = 1e9 + 7;
const int shf = 5e3 + 2;
int bit[maxn];
inline void clr() { fill(bit, bit + maxn, 0); }
inline void upd(int x, int v) {
  while (x < maxn) {
    bit[x] += v;
    x += x & (-x);
  }
}
inline int qry(int x) {
  int ans = 0;
  while (x > 0) {
    ans += bit[x];
    x -= x & (-x);
  }
  return ans;
}
inline int rq(int l, int r) { return qry(r) - qry(l - 1); }
inline bool ix(pair<pair<int, int>, int> h, pair<pair<int, int>, int> v) {
  return (h.first.first <= v.second && v.second <= h.first.second &&
          v.first.first <= h.second && h.second <= v.first.second);
}
vector<pair<pair<int, int>, int> > H, V;
vector<int> Y[maxn];
vector<pair<pair<int, int>, int> > Z;
vector<pair<int, pair<int, int> > > ev;
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
  };
  int n;
  cin >> n;
  for (int j = 0; j < n; j++) {
    int x1, x2, y1, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += shf;
    x2 += shf;
    y1 += shf;
    y2 += shf;
    if (x1 == x2) {
      V.push_back({{min(y1, y2), max(y1, y2)}, x1});
    } else {
      H.push_back({{min(x1, x2), max(x1, x2)}, y1});
      Y[y1].push_back(H.size() - 1);
    }
  }
  long long ans = 0ll;
  for (int j = 0; j < H.size(); j++) {
    ;
    Z.clear();
    ev.clear();
    for (int i = 0; i < V.size(); i++) {
      if (!ix(H[j], V[i])) continue;
      Z.push_back(V[i]);
      ev.push_back({V[i].first.first, {V[i].second, 1}});
      ev.push_back({V[i].first.second + 1, {V[i].second, -1}});
    }
    sort(ev.begin(), ev.end());
    ;
    clr();
    int i = 0;
    for (int y = 0; y < maxn; y++) {
      while (i < ev.size() && ev[i].first == y) {
        upd(ev[i].second.first, ev[i].second.second);
        i++;
      }
      for (int z : Y[y]) {
        if (z <= j) continue;
        int l = H[z].first.first, r = H[z].first.second;
        long long q = rq(l, r);
        ans += q * (q - 1) / 2;
      }
    }
  }
  cout << ans << "\n";
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int grand(int x) { return uniform_int_distribution<int>(0, x - 1)(rng); }
long long gcd(long long x, long long y) {
  if (x < y) return gcd(y, x);
  if (y == 0) return x;
  return gcd(y, x % y);
}
const long long mod = 1e9 + 7;
long long modexp(long long x, long long ex) {
  long long ans = 1ll;
  while (ex > 0) {
    if (ex & 1ll) ans = (ans * x) % mod;
    ex >>= 1ll;
    x = (x * x) % mod;
  }
  return ans;
}
const int maxn = 1e4 + 7;
const long long inf = 1e9 + 7;
const int shf = 5e3 + 2;
int bit[maxn];
inline void clr() { fill(bit, bit + maxn, 0); }
inline void upd(int x, int v) {
  while (x < maxn) {
    bit[x] += v;
    x += x & (-x);
  }
}
inline int qry(int x) {
  int ans = 0;
  while (x > 0) {
    ans += bit[x];
    x -= x & (-x);
  }
  return ans;
}
inline int rq(int l, int r) { return qry(r) - qry(l - 1); }
inline bool ix(pair<pair<int, int>, int> h, pair<pair<int, int>, int> v) {
  return (h.first.first <= v.second && v.second <= h.first.second &&
          v.first.first <= h.second && h.second <= v.first.second);
}
vector<pair<pair<int, int>, int> > H, V;
vector<int> Y[maxn];
vector<pair<pair<int, int>, int> > Z;
vector<pair<int, pair<int, int> > > ev;
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
  };
  int n;
  cin >> n;
  for (int j = 0; j < n; j++) {
    int x1, x2, y1, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += shf;
    x2 += shf;
    y1 += shf;
    y2 += shf;
    if (x1 == x2) {
      V.push_back({{min(y1, y2), max(y1, y2)}, x1});
    } else {
      H.push_back({{min(x1, x2), max(x1, x2)}, y1});
      Y[y1].push_back(H.size() - 1);
    }
  }
  long long ans = 0ll;
  for (int j = 0; j < H.size(); j++) {
    ;
    Z.clear();
    ev.clear();
    for (int i = 0; i < V.size(); i++) {
      if (!ix(H[j], V[i])) continue;
      Z.push_back(V[i]);
      ev.push_back({V[i].first.first, {V[i].second, 1}});
      ev.push_back({V[i].first.second + 1, {V[i].second, -1}});
    }
    sort(ev.begin(), ev.end());
    ;
    clr();
    int i = 0;
    for (int y = 0; y < maxn; y++) {
      while (i < ev.size() && ev[i].first == y) {
        upd(ev[i].second.first, ev[i].second.second);
        i++;
      }
      for (int z : Y[y]) {
        if (z <= j) continue;
        int l = H[z].first.first, r = H[z].first.second;
        long long q = rq(l, r);
        ans += q * (q - 1) / 2;
      }
    }
  }
  cout << ans << "\n";
}

```