#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int mod = 1000000007;
int n;
const int maxn = 5010;
template <class T>
struct BIT {
  T c[maxn];
  BIT() { clear(); }
  inline void clear() { memset(c, 0, sizeof(c)); }
  inline T lowbit(T &x) { return x & (-x); }
  inline void add(int x, T d) {
    while (x <= maxn) {
      c[x] += d;
      x += lowbit(x);
    }
  }
  inline T get(int x) {
    T res = 0;
    while (x) {
      res += c[x];
      x -= lowbit(x);
    }
    return res;
  }
};
BIT<int> bt[5010];
pair<int, pair<int, int> > h[5010], v[5010];
int th, tv;
int hash_x[10010], hash_y[10010], idx[100010], idy[100010];
int tot_x, tot_y;
bool cmp(pair<int, pair<int, int> > A, pair<int, pair<int, int> > B) {
  return A.second.first < B.second.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int a1, b1, a2, b2;
    cin >> a1 >> b1 >> a2 >> b2;
    a1 += 5002, b1 += 5002, a2 += 5002, b2 += 5002;
    if (a1 == a2) {
      if (b1 > b2) swap(b1, b2);
      v[++tv] = make_pair(a1, make_pair(b1, b2));
    } else if (b1 == b2) {
      if (a1 > a2) swap(a1, a2);
      h[++th] = make_pair(b1, make_pair(a1, a2));
      hash_y[++tot_y] = b1;
      hash_x[++tot_x] = a2;
    }
  }
  sort(v + 1, v + tv + 1);
  sort(h + 1, h + th + 1, cmp);
  sort(hash_x + 1, hash_x + tot_x + 1);
  sort(hash_y + 1, hash_y + tot_y + 1);
  tot_x = unique(hash_x + 1, hash_x + tot_x + 1) - hash_x - 1;
  tot_y = unique(hash_y + 1, hash_y + tot_y + 1) - hash_y - 1;
  for (int i = 1; i <= tot_x; ++i) idx[hash_x[i]] = i;
  for (int i = 1; i <= tot_y; ++i) idy[hash_y[i]] = i;
  int p = 0;
  long long ans = 0;
  for (int i = 1; i <= tv; ++i) {
    while (p + 1 <= th && h[p + 1].second.first <= v[i].first) {
      int id = idx[h[p + 1].second.second];
      for (int k = 1; k <= id; ++k) bt[k].add(idy[h[p + 1].first], 1);
      p++;
    }
    for (int j = i + 1; j <= tv; ++j)
      if (v[j].first > v[i].first) {
        int ydown = max(v[i].second.first, v[j].second.first);
        int yup = min(v[i].second.second, v[j].second.second);
        if (ydown <= yup) {
          int pdown =
              lower_bound(hash_y + 1, hash_y + tot_y + 1, ydown) - hash_y;
          int pup =
              upper_bound(hash_y + 1, hash_y + tot_y + 1, yup) - hash_y - 1;
          if (pdown <= pup) {
            int xdown =
                lower_bound(hash_x + 1, hash_x + tot_x + 1, v[j].first) -
                hash_x;
            int cnt = bt[xdown].get(pup) - bt[xdown].get(pdown - 1);
            if (cnt > 0) ans += 1LL * cnt * (cnt - 1) / 2;
          }
        }
      }
  }
  cout << ans << endl;
  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")
using namespace std;
const int mod = 1000000007;
int n;
const int maxn = 5010;
template <class T>
struct BIT {
  T c[maxn];
  BIT() { clear(); }
  inline void clear() { memset(c, 0, sizeof(c)); }
  inline T lowbit(T &x) { return x & (-x); }
  inline void add(int x, T d) {
    while (x <= maxn) {
      c[x] += d;
      x += lowbit(x);
    }
  }
  inline T get(int x) {
    T res = 0;
    while (x) {
      res += c[x];
      x -= lowbit(x);
    }
    return res;
  }
};
BIT<int> bt[5010];
pair<int, pair<int, int> > h[5010], v[5010];
int th, tv;
int hash_x[10010], hash_y[10010], idx[100010], idy[100010];
int tot_x, tot_y;
bool cmp(pair<int, pair<int, int> > A, pair<int, pair<int, int> > B) {
  return A.second.first < B.second.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int a1, b1, a2, b2;
    cin >> a1 >> b1 >> a2 >> b2;
    a1 += 5002, b1 += 5002, a2 += 5002, b2 += 5002;
    if (a1 == a2) {
      if (b1 > b2) swap(b1, b2);
      v[++tv] = make_pair(a1, make_pair(b1, b2));
    } else if (b1 == b2) {
      if (a1 > a2) swap(a1, a2);
      h[++th] = make_pair(b1, make_pair(a1, a2));
      hash_y[++tot_y] = b1;
      hash_x[++tot_x] = a2;
    }
  }
  sort(v + 1, v + tv + 1);
  sort(h + 1, h + th + 1, cmp);
  sort(hash_x + 1, hash_x + tot_x + 1);
  sort(hash_y + 1, hash_y + tot_y + 1);
  tot_x = unique(hash_x + 1, hash_x + tot_x + 1) - hash_x - 1;
  tot_y = unique(hash_y + 1, hash_y + tot_y + 1) - hash_y - 1;
  for (int i = 1; i <= tot_x; ++i) idx[hash_x[i]] = i;
  for (int i = 1; i <= tot_y; ++i) idy[hash_y[i]] = i;
  int p = 0;
  long long ans = 0;
  for (int i = 1; i <= tv; ++i) {
    while (p + 1 <= th && h[p + 1].second.first <= v[i].first) {
      int id = idx[h[p + 1].second.second];
      for (int k = 1; k <= id; ++k) bt[k].add(idy[h[p + 1].first], 1);
      p++;
    }
    for (int j = i + 1; j <= tv; ++j)
      if (v[j].first > v[i].first) {
        int ydown = max(v[i].second.first, v[j].second.first);
        int yup = min(v[i].second.second, v[j].second.second);
        if (ydown <= yup) {
          int pdown =
              lower_bound(hash_y + 1, hash_y + tot_y + 1, ydown) - hash_y;
          int pup =
              upper_bound(hash_y + 1, hash_y + tot_y + 1, yup) - hash_y - 1;
          if (pdown <= pup) {
            int xdown =
                lower_bound(hash_x + 1, hash_x + tot_x + 1, v[j].first) -
                hash_x;
            int cnt = bt[xdown].get(pup) - bt[xdown].get(pdown - 1);
            if (cnt > 0) ans += 1LL * cnt * (cnt - 1) / 2;
          }
        }
      }
  }
  cout << ans << endl;
  return 0;
}

```