#include <bits/stdc++.h>
using namespace std;
const int maxn = 10009;
const int D = 5003;
const int INF = 1000000000;
struct BIT {
  int c[maxn + 100], n;
  void init(int n) {
    this->n = n;
    memset(c, 0, sizeof(c));
  }
  int sum(int x) {
    int ans = 0;
    for (; x; x -= x & -x) ans += c[x];
    return ans;
  }
  void add(int x, int y) {
    for (; x <= n; x += x & -x) c[x] += y;
  }
} b;
int n;
vector<pair<int, int> > vs[maxn], hs[maxn];
vector<int> d[maxn];
int get(int L, int R) { return b.sum(R) - b.sum(L - 1); }
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  b.init(maxn);
  int x1, x2, y1, y2;
  for (int i = 0; i < n; i++) {
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += D, x2 += D, y1 += D, y2 += D;
    if (y1 == y2) {
      if (x1 > x2) swap(x1, x2);
      hs[y1].push_back(make_pair(x1, x2));
    } else {
      if (y1 > y2) swap(y1, y2);
      vs[x1].push_back(make_pair(y1, y2));
    }
  }
  long long ans = 0;
  for (int y = 0; y < maxn; y++) {
    for (auto s : hs[y]) {
      for (int i = 0; i < maxn; i++) d[i].clear();
      int L = s.first, R = s.second;
      for (int x = L; x <= R; x++)
        for (auto s2 : vs[x]) {
          if (s2.first <= y && s2.second > y) {
            d[s2.second].push_back(x);
            b.add(x, 1);
          }
        }
      for (int y2 = y + 1; y2 < maxn; y2++) {
        for (auto s2 : hs[y2]) {
          long long s = get(s2.first, s2.second);
          ans += (s - 1) * s / 2;
        }
        for (auto x : d[y2]) b.add(x, -1);
      }
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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 maxn = 10009;
const int D = 5003;
const int INF = 1000000000;
struct BIT {
  int c[maxn + 100], n;
  void init(int n) {
    this->n = n;
    memset(c, 0, sizeof(c));
  }
  int sum(int x) {
    int ans = 0;
    for (; x; x -= x & -x) ans += c[x];
    return ans;
  }
  void add(int x, int y) {
    for (; x <= n; x += x & -x) c[x] += y;
  }
} b;
int n;
vector<pair<int, int> > vs[maxn], hs[maxn];
vector<int> d[maxn];
int get(int L, int R) { return b.sum(R) - b.sum(L - 1); }
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  b.init(maxn);
  int x1, x2, y1, y2;
  for (int i = 0; i < n; i++) {
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += D, x2 += D, y1 += D, y2 += D;
    if (y1 == y2) {
      if (x1 > x2) swap(x1, x2);
      hs[y1].push_back(make_pair(x1, x2));
    } else {
      if (y1 > y2) swap(y1, y2);
      vs[x1].push_back(make_pair(y1, y2));
    }
  }
  long long ans = 0;
  for (int y = 0; y < maxn; y++) {
    for (auto s : hs[y]) {
      for (int i = 0; i < maxn; i++) d[i].clear();
      int L = s.first, R = s.second;
      for (int x = L; x <= R; x++)
        for (auto s2 : vs[x]) {
          if (s2.first <= y && s2.second > y) {
            d[s2.second].push_back(x);
            b.add(x, 1);
          }
        }
      for (int y2 = y + 1; y2 < maxn; y2++) {
        for (auto s2 : hs[y2]) {
          long long s = get(s2.first, s2.second);
          ans += (s - 1) * s / 2;
        }
        for (auto x : d[y2]) b.add(x, -1);
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```