#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1000011;
const long long INF = (1 << 29) + 123;
const long long MOD = 1000000007;
const long double PI = 4 * atan((long double)1);
const int N = 10001;
int n;
const int bias = N / 2;
vector<pair<int, int> > h[N], v[N];
int fenw[N + 1];
vector<int> curY[N];
void upd(int p, int v) {
  p++;
  while (p <= N) {
    fenw[p] += v;
    p += (p & -p);
  }
}
int query(int p) {
  int sum = 0;
  p++;
  while (p > 0) {
    sum += fenw[p];
    p -= (p & -p);
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int x1, y1, x2, y2;
  for (int i = 0; i < n; i++) {
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += bias;
    y1 += bias;
    x2 += bias;
    y2 += bias;
    if (x1 == x2) {
      v[x1].push_back(make_pair(min(y1, y2), max(y1, y2)));
    } else {
      h[y1].push_back(make_pair(min(x1, x2), max(x1, x2)));
    }
  }
  long long ans = 0;
  for (int y = 0; y < N; y++) {
    for (auto& a : h[y]) {
      memset(fenw, 0, sizeof(fenw));
      for (int i = 0; i < N; i++) curY[i].clear();
      for (int x = a.first; x < a.second + 1; x++) {
        for (auto& ax : v[x]) {
          if (ax.first <= y && ax.second > y) {
            upd(x, 1);
            curY[ax.second].push_back(x);
          }
        }
      }
      for (int yy = y + 1; yy < N; yy++) {
        for (auto& a1 : h[yy]) {
          long long temp = query(a1.second) - query(a1.first - 1);
          ans += temp * (temp - 1) / 2;
        }
        for (auto& a1 : curY[yy]) upd(a1, -1);
      }
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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;
const int MAX_N = 1000011;
const long long INF = (1 << 29) + 123;
const long long MOD = 1000000007;
const long double PI = 4 * atan((long double)1);
const int N = 10001;
int n;
const int bias = N / 2;
vector<pair<int, int> > h[N], v[N];
int fenw[N + 1];
vector<int> curY[N];
void upd(int p, int v) {
  p++;
  while (p <= N) {
    fenw[p] += v;
    p += (p & -p);
  }
}
int query(int p) {
  int sum = 0;
  p++;
  while (p > 0) {
    sum += fenw[p];
    p -= (p & -p);
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int x1, y1, x2, y2;
  for (int i = 0; i < n; i++) {
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += bias;
    y1 += bias;
    x2 += bias;
    y2 += bias;
    if (x1 == x2) {
      v[x1].push_back(make_pair(min(y1, y2), max(y1, y2)));
    } else {
      h[y1].push_back(make_pair(min(x1, x2), max(x1, x2)));
    }
  }
  long long ans = 0;
  for (int y = 0; y < N; y++) {
    for (auto& a : h[y]) {
      memset(fenw, 0, sizeof(fenw));
      for (int i = 0; i < N; i++) curY[i].clear();
      for (int x = a.first; x < a.second + 1; x++) {
        for (auto& ax : v[x]) {
          if (ax.first <= y && ax.second > y) {
            upd(x, 1);
            curY[ax.second].push_back(x);
          }
        }
      }
      for (int yy = y + 1; yy < N; yy++) {
        for (auto& a1 : h[yy]) {
          long long temp = query(a1.second) - query(a1.first - 1);
          ans += temp * (temp - 1) / 2;
        }
        for (auto& a1 : curY[yy]) upd(a1, -1);
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```