#include <bits/stdc++.h>
using namespace std;
using INT = long long;
using pii = pair<int, int>;
const int NN = 10101;
namespace Bit {
int B[NN << 1];
void clear() { memset(B, 0, sizeof B); }
void update(int u, int v) {
  for (; u < NN; u += u & -u) B[u] += v;
}
int calc(int u, int ans = 0) {
  for (; u; u ^= u & -u) ans += B[u];
  return ans;
}
};  // namespace Bit
using namespace Bit;
vector<pii> row[NN], col[NN], rev[NN];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int first, second, xx, yy;
    scanf("%d %d %d %d", &first, &second, &xx, &yy);
    first += 5001;
    second += 5001;
    xx += 5001;
    yy += 5001;
    if (first == xx) {
      col[first].push_back(pii(min(second, yy), max(second, yy)));
      rev[max(second, yy)].push_back(pii(min(second, yy), first));
    } else {
      row[second].push_back(pii(min(first, xx), max(first, xx)));
    }
  }
  INT ans = 0;
  for (int i = 0; i < NN; i++) {
    for (auto p : row[i]) {
      clear();
      for (int j = p.first; j <= p.second; j++) {
        for (auto q : col[j]) {
          if (q.first <= i && q.second > i) {
            update(j, 1);
            break;
          }
        }
      }
      for (int j = i + 1; j < NN; j++) {
        for (auto q : row[j]) {
          int st = max(q.first, p.first);
          int ed = min(q.second, p.second);
          if (st > ed) continue;
          int cnt = calc(ed) - calc(st - 1);
          ans += 1LL * cnt * (cnt - 1) / 2;
        }
        for (auto q : rev[j]) {
          int u = q.second;
          int v = q.first;
          if (v > i) continue;
          update(u, -1);
        }
      }
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in cpp 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;
using INT = long long;
using pii = pair<int, int>;
const int NN = 10101;
namespace Bit {
int B[NN << 1];
void clear() { memset(B, 0, sizeof B); }
void update(int u, int v) {
  for (; u < NN; u += u & -u) B[u] += v;
}
int calc(int u, int ans = 0) {
  for (; u; u ^= u & -u) ans += B[u];
  return ans;
}
};  // namespace Bit
using namespace Bit;
vector<pii> row[NN], col[NN], rev[NN];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int first, second, xx, yy;
    scanf("%d %d %d %d", &first, &second, &xx, &yy);
    first += 5001;
    second += 5001;
    xx += 5001;
    yy += 5001;
    if (first == xx) {
      col[first].push_back(pii(min(second, yy), max(second, yy)));
      rev[max(second, yy)].push_back(pii(min(second, yy), first));
    } else {
      row[second].push_back(pii(min(first, xx), max(first, xx)));
    }
  }
  INT ans = 0;
  for (int i = 0; i < NN; i++) {
    for (auto p : row[i]) {
      clear();
      for (int j = p.first; j <= p.second; j++) {
        for (auto q : col[j]) {
          if (q.first <= i && q.second > i) {
            update(j, 1);
            break;
          }
        }
      }
      for (int j = i + 1; j < NN; j++) {
        for (auto q : row[j]) {
          int st = max(q.first, p.first);
          int ed = min(q.second, p.second);
          if (st > ed) continue;
          int cnt = calc(ed) - calc(st - 1);
          ans += 1LL * cnt * (cnt - 1) / 2;
        }
        for (auto q : rev[j]) {
          int u = q.second;
          int v = q.first;
          if (v > i) continue;
          update(u, -1);
        }
      }
    }
  }
  cout << ans;
  return 0;
}

```