#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 9, crange = 5001;
vector<pair<int, pair<int, int>>> ver, hor;
vector<int> edline[N];
int f[N];
void update(int idx, int val) {
  for (; idx < N; idx += idx & -idx) {
    f[idx] += val;
  }
}
int getsum(int idx) {
  int cnt = 0;
  for (; idx; idx -= idx & -idx) {
    cnt += f[idx];
  }
  return cnt;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1, x1, y1, x2, y2; i <= n; i++) {
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += crange;
    x2 += crange;
    y1 += crange;
    y2 += crange;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2) {
      hor.push_back({x1, {y1, y2}});
    } else
      ver.push_back({y1, {x1, x2}});
  }
  sort(hor.begin(), hor.end());
  long long ans = 0;
  for (int i = 0, x, y1, y2; i < hor.size(); i++) {
    x = hor[i].first;
    y1 = hor[i].second.first;
    y2 = hor[i].second.second;
    for (auto it : ver) {
      if (it.second.first <= x && x < it.second.second) {
        update(it.first, 1);
        edline[it.second.second].push_back(it.first);
      }
    }
    int r = x;
    for (int j = i + 1; j < hor.size(); j++) {
      while (r < hor[j].first) {
        while (!edline[r].empty()) {
          update(edline[r].back(), -1);
          edline[r].pop_back();
        }
        r++;
      }
      int R = min(y2, hor[j].second.second), L = max(y1, hor[j].second.first);
      if (L <= R) {
        int cnt = getsum(R) - getsum(L - 1);
        ans += cnt * (cnt - 1) / 2;
      }
    }
    while (r < N) {
      while (!edline[r].empty()) {
        update(edline[r].back(), -1);
        edline[r].pop_back();
      }
      r++;
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
const int N = 1e4 + 9, crange = 5001;
vector<pair<int, pair<int, int>>> ver, hor;
vector<int> edline[N];
int f[N];
void update(int idx, int val) {
  for (; idx < N; idx += idx & -idx) {
    f[idx] += val;
  }
}
int getsum(int idx) {
  int cnt = 0;
  for (; idx; idx -= idx & -idx) {
    cnt += f[idx];
  }
  return cnt;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1, x1, y1, x2, y2; i <= n; i++) {
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += crange;
    x2 += crange;
    y1 += crange;
    y2 += crange;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2) {
      hor.push_back({x1, {y1, y2}});
    } else
      ver.push_back({y1, {x1, x2}});
  }
  sort(hor.begin(), hor.end());
  long long ans = 0;
  for (int i = 0, x, y1, y2; i < hor.size(); i++) {
    x = hor[i].first;
    y1 = hor[i].second.first;
    y2 = hor[i].second.second;
    for (auto it : ver) {
      if (it.second.first <= x && x < it.second.second) {
        update(it.first, 1);
        edline[it.second.second].push_back(it.first);
      }
    }
    int r = x;
    for (int j = i + 1; j < hor.size(); j++) {
      while (r < hor[j].first) {
        while (!edline[r].empty()) {
          update(edline[r].back(), -1);
          edline[r].pop_back();
        }
        r++;
      }
      int R = min(y2, hor[j].second.second), L = max(y1, hor[j].second.first);
      if (L <= R) {
        int cnt = getsum(R) - getsum(L - 1);
        ans += cnt * (cnt - 1) / 2;
      }
    }
    while (r < N) {
      while (!edline[r].empty()) {
        update(edline[r].back(), -1);
        edline[r].pop_back();
      }
      r++;
    }
  }
  cout << ans << '\n';
  return 0;
}

```