#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
const int mxN = 1e4 + 5;
template <class T>
T gcd(T a, T b) {
  return ((b == 0) ? a : gcd(b, a % b));
}
vector<int> del[mxN];
vector<array<int, 2>> hor[mxN];
template <class T>
struct BIT {
  vector<T> bit;
  BIT() { bit.assign(mxN, 0); }
  void update(int ind, T delta) {
    for (; ind < mxN; ind = ind | (ind + 1)) {
      bit[ind] += delta;
    }
  }
  T query(int ind) {
    T sum = 0;
    for (; ind >= 0; ind = (ind & (ind + 1)) - 1) {
      sum += bit[ind];
    }
    return sum;
  }
  T query(int l, int r) { return query(r) - (l ? query(l - 1) : 0); }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<array<int, 3>> ver;
  for (int i = 0; i < (int)(n); i++) {
    int x, y, x2, y2;
    cin >> x >> y >> x2 >> y2;
    x += 5000, y += 5000;
    x2 += 5000, y2 += 5000;
    if (x2 < x) swap(x, x2);
    if (y2 < y) swap(y, y2);
    if (x == x2)
      ver.push_back({x, y, y2});
    else
      hor[x].push_back({y, x2});
  }
  sort(ver.begin(), ver.end());
  long long ans = 0LL;
  for (int i = 0; i < (int)(ver.size()); i++) {
    BIT<int> bit;
    for (int j = 0; j < (int)(ver[i][0] + 1); j++) {
      for (array<int, 2> x : hor[j])
        if (x[1] > ver[i][0]) {
          bit.update(x[0], 1);
          del[x[1]].push_back(x[0]);
        }
    }
    for (int j = (int)(i + 1); j < (int)(ver.size()); j++) {
      int mn = max(ver[i][1], ver[j][1]);
      int mx = min(ver[i][2], ver[j][2]);
      for (int k = (int)(ver[j - 1][0]); k < (int)(ver[j][0]); k++) {
        while (del[k].size()) {
          bit.update(del[k].back(), -1);
          del[k].pop_back();
        }
      }
      if (mn > mx) continue;
      int nm = bit.query(mn, mx);
      ans += (long long)(1LL * nm * (nm - 1) / 2);
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 optimize("O3")
#pragma GCC target("sse4")
using namespace std;
const int mxN = 1e4 + 5;
template <class T>
T gcd(T a, T b) {
  return ((b == 0) ? a : gcd(b, a % b));
}
vector<int> del[mxN];
vector<array<int, 2>> hor[mxN];
template <class T>
struct BIT {
  vector<T> bit;
  BIT() { bit.assign(mxN, 0); }
  void update(int ind, T delta) {
    for (; ind < mxN; ind = ind | (ind + 1)) {
      bit[ind] += delta;
    }
  }
  T query(int ind) {
    T sum = 0;
    for (; ind >= 0; ind = (ind & (ind + 1)) - 1) {
      sum += bit[ind];
    }
    return sum;
  }
  T query(int l, int r) { return query(r) - (l ? query(l - 1) : 0); }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<array<int, 3>> ver;
  for (int i = 0; i < (int)(n); i++) {
    int x, y, x2, y2;
    cin >> x >> y >> x2 >> y2;
    x += 5000, y += 5000;
    x2 += 5000, y2 += 5000;
    if (x2 < x) swap(x, x2);
    if (y2 < y) swap(y, y2);
    if (x == x2)
      ver.push_back({x, y, y2});
    else
      hor[x].push_back({y, x2});
  }
  sort(ver.begin(), ver.end());
  long long ans = 0LL;
  for (int i = 0; i < (int)(ver.size()); i++) {
    BIT<int> bit;
    for (int j = 0; j < (int)(ver[i][0] + 1); j++) {
      for (array<int, 2> x : hor[j])
        if (x[1] > ver[i][0]) {
          bit.update(x[0], 1);
          del[x[1]].push_back(x[0]);
        }
    }
    for (int j = (int)(i + 1); j < (int)(ver.size()); j++) {
      int mn = max(ver[i][1], ver[j][1]);
      int mx = min(ver[i][2], ver[j][2]);
      for (int k = (int)(ver[j - 1][0]); k < (int)(ver[j][0]); k++) {
        while (del[k].size()) {
          bit.update(del[k].back(), -1);
          del[k].pop_back();
        }
      }
      if (mn > mx) continue;
      int nm = bit.query(mn, mx);
      ans += (long long)(1LL * nm * (nm - 1) / 2);
    }
  }
  cout << ans << '\n';
  return 0;
}

```