#include <bits/stdc++.h>
using namespace std;
constexpr int MOD = 1e9 + 7;
constexpr int MAXN = 1e4 + 10;
using ll = long long;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;
ll N, M, K, T;
vector<Pii> segs[MAXN];
vector<Pii> hor[MAXN], ver[MAXN];
vector<int> d[MAXN];
int f[MAXN];
constexpr int D = 5001;
void Upd(int pos, int x) {
  while (pos <= MAXN) {
    f[pos] += x;
    pos |= pos + 1;
  }
}
int Get(int pos) {
  int ret = 0;
  while (pos >= 0) {
    ret += f[pos];
    pos = (pos & (pos + 1)) - 1;
  }
  return ret;
}
int Get(int l, int r) { return Get(r) - Get(l - 1); }
ll Solution() {
  for (int i = (int)(0); i < (int)(N); ++i) {
    int x1 = segs[i][0].first, y1 = segs[i][0].second;
    int x2 = segs[i][1].first, y2 = segs[i][1].second;
    x1 += D, y1 += D, x2 += D, y2 += D;
    if (y1 == y2)
      hor[y1].emplace_back(min(x1, x2), max(x1, x2));
    else
      ver[x1].emplace_back(min(y1, y2), max(y1, y2));
  }
  ll ans = 0;
  for (int y = (int)(0); y < (int)(MAXN); ++y) {
    for (auto s : hor[y]) {
      for (int i = (int)(0); i < (int)(MAXN); ++i) d[i].clear();
      memset(f, 0, sizeof(f));
      int l = s.first, r = s.second;
      for (int x = (int)(l); x < (int)(r + 1); ++x) {
        for (auto s2 : ver[x]) {
          if (s2.first <= y && s2.second > y) {
            d[s2.second].emplace_back(x);
            Upd(x, 1);
          }
        }
      }
      for (int y2 = (int)(y + 1); y2 < (int)(MAXN); ++y2) {
        for (auto s2 : hor[y2]) {
          ll cur = Get(s2.first, s2.second);
          ans += cur * (cur - 1) / 2;
        }
        for (auto x : d[y2]) Upd(x, -1);
      }
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  while (cin >> N) {
    int x1, y1, x2, y2;
    for (int i = (int)(0); i < (int)(N); ++i) {
      cin >> x1 >> y1 >> x2 >> y2;
      segs[i].emplace_back(x1, y1);
      segs[i].emplace_back(x2, y2);
    }
    cout << Solution() << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
constexpr int MOD = 1e9 + 7;
constexpr int MAXN = 1e4 + 10;
using ll = long long;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;
ll N, M, K, T;
vector<Pii> segs[MAXN];
vector<Pii> hor[MAXN], ver[MAXN];
vector<int> d[MAXN];
int f[MAXN];
constexpr int D = 5001;
void Upd(int pos, int x) {
  while (pos <= MAXN) {
    f[pos] += x;
    pos |= pos + 1;
  }
}
int Get(int pos) {
  int ret = 0;
  while (pos >= 0) {
    ret += f[pos];
    pos = (pos & (pos + 1)) - 1;
  }
  return ret;
}
int Get(int l, int r) { return Get(r) - Get(l - 1); }
ll Solution() {
  for (int i = (int)(0); i < (int)(N); ++i) {
    int x1 = segs[i][0].first, y1 = segs[i][0].second;
    int x2 = segs[i][1].first, y2 = segs[i][1].second;
    x1 += D, y1 += D, x2 += D, y2 += D;
    if (y1 == y2)
      hor[y1].emplace_back(min(x1, x2), max(x1, x2));
    else
      ver[x1].emplace_back(min(y1, y2), max(y1, y2));
  }
  ll ans = 0;
  for (int y = (int)(0); y < (int)(MAXN); ++y) {
    for (auto s : hor[y]) {
      for (int i = (int)(0); i < (int)(MAXN); ++i) d[i].clear();
      memset(f, 0, sizeof(f));
      int l = s.first, r = s.second;
      for (int x = (int)(l); x < (int)(r + 1); ++x) {
        for (auto s2 : ver[x]) {
          if (s2.first <= y && s2.second > y) {
            d[s2.second].emplace_back(x);
            Upd(x, 1);
          }
        }
      }
      for (int y2 = (int)(y + 1); y2 < (int)(MAXN); ++y2) {
        for (auto s2 : hor[y2]) {
          ll cur = Get(s2.first, s2.second);
          ans += cur * (cur - 1) / 2;
        }
        for (auto x : d[y2]) Upd(x, -1);
      }
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  while (cin >> N) {
    int x1, y1, x2, y2;
    for (int i = (int)(0); i < (int)(N); ++i) {
      cin >> x1 >> y1 >> x2 >> y2;
      segs[i].emplace_back(x1, y1);
      segs[i].emplace_back(x2, y2);
    }
    cout << Solution() << '\n';
  }
  return 0;
}

```