#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007LL;
long long mod2 = 998244353LL;
int n;
vector<pair<int, pair<int, int>>> hs;
vector<pair<int, pair<int, int>>> vs;
int BIT[5005 * 3];
int lowbit(int x) { return x & (-x); }
void update(int pos, int val) {
  while (pos <= 10002) {
    BIT[pos] += val;
    pos += lowbit(pos);
  }
}
int query(int pos) {
  int res = 0;
  while (pos > 0) {
    res += BIT[pos];
    pos -= lowbit(pos);
  }
  return res;
}
int main() {
  cin >> n;
  hs.clear();
  vs.clear();
  for (int i = 0; i < n; ++i) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    if (a == c) {
      vs.push_back(make_pair(a, make_pair(max(b, d), min(b, d))));
    } else {
      hs.push_back(make_pair(b, make_pair(min(a, c), max(a, c))));
    }
  }
  sort(vs.begin(), vs.end());
  sort(hs.begin(), hs.end());
  long long ans = 0LL;
  reverse(hs.begin(), hs.end());
  for (int i = 0; i < hs.size(); ++i) {
    memset(BIT, 0, sizeof(BIT));
    vector<pair<int, int>> cdd;
    for (int j = 0; j < vs.size(); ++j) {
      int a = vs[j].second.first;
      int b = vs[j].second.second;
      if (a >= hs[i].first && b <= hs[i].first) {
        cdd.push_back(make_pair(b, vs[j].first));
      }
    }
    sort(cdd.begin(), cdd.end());
    reverse(cdd.begin(), cdd.end());
    for (int j = 0; j < cdd.size(); ++j) {
      update(cdd[j].second + 5000 + 1, 1);
    }
    int k = 0;
    for (int j = i + 1; j < hs.size(); ++j) {
      if (hs[j].first == hs[i].first) {
        continue;
      }
      int a = max(hs[i].second.first, hs[j].second.first);
      int b = min(hs[i].second.second, hs[j].second.second);
      if (a >= b) continue;
      while (k < cdd.size() && cdd[k].first > hs[j].first) {
        update(cdd[k].second + 5000 + 1, -1);
        k++;
      }
      int tot = query(b + 5000 + 1) - query(a + 5000);
      if (tot >= 1) {
        ans += (long long)tot * (long long)(tot - 1) / 2;
      }
    }
  }
  printf("%lld\n", ans);
  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;
long long mod = 1000000007LL;
long long mod2 = 998244353LL;
int n;
vector<pair<int, pair<int, int>>> hs;
vector<pair<int, pair<int, int>>> vs;
int BIT[5005 * 3];
int lowbit(int x) { return x & (-x); }
void update(int pos, int val) {
  while (pos <= 10002) {
    BIT[pos] += val;
    pos += lowbit(pos);
  }
}
int query(int pos) {
  int res = 0;
  while (pos > 0) {
    res += BIT[pos];
    pos -= lowbit(pos);
  }
  return res;
}
int main() {
  cin >> n;
  hs.clear();
  vs.clear();
  for (int i = 0; i < n; ++i) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    if (a == c) {
      vs.push_back(make_pair(a, make_pair(max(b, d), min(b, d))));
    } else {
      hs.push_back(make_pair(b, make_pair(min(a, c), max(a, c))));
    }
  }
  sort(vs.begin(), vs.end());
  sort(hs.begin(), hs.end());
  long long ans = 0LL;
  reverse(hs.begin(), hs.end());
  for (int i = 0; i < hs.size(); ++i) {
    memset(BIT, 0, sizeof(BIT));
    vector<pair<int, int>> cdd;
    for (int j = 0; j < vs.size(); ++j) {
      int a = vs[j].second.first;
      int b = vs[j].second.second;
      if (a >= hs[i].first && b <= hs[i].first) {
        cdd.push_back(make_pair(b, vs[j].first));
      }
    }
    sort(cdd.begin(), cdd.end());
    reverse(cdd.begin(), cdd.end());
    for (int j = 0; j < cdd.size(); ++j) {
      update(cdd[j].second + 5000 + 1, 1);
    }
    int k = 0;
    for (int j = i + 1; j < hs.size(); ++j) {
      if (hs[j].first == hs[i].first) {
        continue;
      }
      int a = max(hs[i].second.first, hs[j].second.first);
      int b = min(hs[i].second.second, hs[j].second.second);
      if (a >= b) continue;
      while (k < cdd.size() && cdd[k].first > hs[j].first) {
        update(cdd[k].second + 5000 + 1, -1);
        k++;
      }
      int tot = query(b + 5000 + 1) - query(a + 5000);
      if (tot >= 1) {
        ans += (long long)tot * (long long)(tot - 1) / 2;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```