#include <bits/stdc++.h>
using namespace std;
const int off = 5001, N = 10009;
vector<pair<int, int>> vs[N], hs[N];
vector<int> open[N];
int bit[N];
int sum(int id) {
  int ans = 0;
  while (id > 0) ans += bit[id], id -= id & -id;
  return ans;
}
void update(int id, int val) {
  while (id <= N) bit[id] += val, id += id & -id;
}
int query(int l, int r) { return sum(r) - sum(l - 1); }
void __sol() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x1, x2, y1, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += off, x2 += off, y1 += off, y2 += off;
    if (y1 == y2)
      hs[y1].push_back({min(x1, x2), max(x1, x2)});
    else
      vs[x1].push_back({min(y1, y2), max(y1, y2)});
  }
  long long ans = 0;
  for (int fix_y = 1; fix_y < N; fix_y++)
    for (auto &s1 : hs[fix_y]) {
      memset(bit, 0, sizeof(bit));
      int l = s1.first, r = s1.second;
      for (int x = l; x <= r; x++)
        for (auto &s2 : vs[x])
          if (s2.first <= fix_y && fix_y < s2.second) {
            open[s2.second].push_back(x);
            update(x, 1);
          }
      for (int y = fix_y + 1; y < N; y++) {
        for (auto &s2 : hs[y]) {
          int curr = query(s2.first, s2.second);
          ans += curr * (curr - 1) / 2;
        }
        for (auto &x : open[y]) update(x, -1);
        open[y].clear();
      }
    }
  cout << ans << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int tc = 1;
  while (tc--) __sol();
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 off = 5001, N = 10009;
vector<pair<int, int>> vs[N], hs[N];
vector<int> open[N];
int bit[N];
int sum(int id) {
  int ans = 0;
  while (id > 0) ans += bit[id], id -= id & -id;
  return ans;
}
void update(int id, int val) {
  while (id <= N) bit[id] += val, id += id & -id;
}
int query(int l, int r) { return sum(r) - sum(l - 1); }
void __sol() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x1, x2, y1, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += off, x2 += off, y1 += off, y2 += off;
    if (y1 == y2)
      hs[y1].push_back({min(x1, x2), max(x1, x2)});
    else
      vs[x1].push_back({min(y1, y2), max(y1, y2)});
  }
  long long ans = 0;
  for (int fix_y = 1; fix_y < N; fix_y++)
    for (auto &s1 : hs[fix_y]) {
      memset(bit, 0, sizeof(bit));
      int l = s1.first, r = s1.second;
      for (int x = l; x <= r; x++)
        for (auto &s2 : vs[x])
          if (s2.first <= fix_y && fix_y < s2.second) {
            open[s2.second].push_back(x);
            update(x, 1);
          }
      for (int y = fix_y + 1; y < N; y++) {
        for (auto &s2 : hs[y]) {
          int curr = query(s2.first, s2.second);
          ans += curr * (curr - 1) / 2;
        }
        for (auto &x : open[y]) update(x, -1);
        open[y].clear();
      }
    }
  cout << ans << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int tc = 1;
  while (tc--) __sol();
  return 0;
}

```