#include <bits/stdc++.h>
using namespace std;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
const int N = 5e3 + 10;
const long long MOD = 1e9 + 7;
const long long M1 = 1e18 + 3, M2 = 1e9 + 7;
const long long p1 = 197, p2 = 229;
const int rx[4] = {-1, 0, 1, 0};
const int ry[4] = {0, 1, 0, -1};
const int hx[8] = {-2, -2, -1, 1, 2, 2, 1, -1};
const int hy[8] = {-1, 1, 2, 2, 1, -1, -2, -2};
const long double eps = 1e-7;
const long double pi = acos(-1.0);
const long long INF = 1e9;
struct cdst {
  int x1, y1, x2, y2;
};
struct fentree {
  vector<int> t;
  fentree() { t.resize(10010); }
  void add(int pos, int x) {
    for (; pos < 10010; pos += (pos & (-pos))) {
      t[pos] += x;
    }
  }
  int sum(int pos) {
    int res = 0;
    for (; pos > 0; pos -= (pos & (-pos))) {
      res += t[pos];
    }
    return res;
  }
  int sum(int l, int r) { return sum(r) - sum(l - 1); }
};
bool comp(cdst a, cdst b) {
  if (a.y2 == b.y2) {
    return a.x1 != a.x2;
  } else
    return a.y2 < b.y2;
}
cdst cs[N];
bool napr[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> cs[i].x1 >> cs[i].y1 >> cs[i].x2 >> cs[i].y2;
    cs[i].x1 += 5001;
    cs[i].y1 += 5001;
    cs[i].x2 += 5001;
    cs[i].y2 += 5001;
    if (cs[i].y1 > cs[i].y2) swap(cs[i].y1, cs[i].y2);
    if (cs[i].x1 > cs[i].x2) swap(cs[i].x1, cs[i].x2);
  }
  sort(cs, cs + n, comp);
  for (int i = 0; i < n; i++) {
    if (cs[i].x1 == cs[i].x2) napr[i] = true;
  }
  fentree ft;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (napr[i]) continue;
    for (int j = 0; j < n; j++) {
      if (napr[j] && cs[i].x1 <= cs[j].x1 && cs[i].x2 >= cs[j].x1 &&
          cs[j].y1 <= cs[i].y1) {
        ft.add(cs[j].x1, 1);
      }
    }
    for (int j = 0; j < n; j++) {
      if (napr[j] && cs[i].x1 <= cs[j].x1 && cs[i].x2 >= cs[j].x1 &&
          cs[j].y1 <= cs[i].y1) {
        ft.add(cs[j].x1, -1);
      } else if (!napr[j] && j > i) {
        long long s = ft.sum(cs[j].x1, cs[j].x2);
        ans += (s * (s - 1)) / 2;
      }
    }
  }
  cout << ans << '\n';
  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;
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
const int N = 5e3 + 10;
const long long MOD = 1e9 + 7;
const long long M1 = 1e18 + 3, M2 = 1e9 + 7;
const long long p1 = 197, p2 = 229;
const int rx[4] = {-1, 0, 1, 0};
const int ry[4] = {0, 1, 0, -1};
const int hx[8] = {-2, -2, -1, 1, 2, 2, 1, -1};
const int hy[8] = {-1, 1, 2, 2, 1, -1, -2, -2};
const long double eps = 1e-7;
const long double pi = acos(-1.0);
const long long INF = 1e9;
struct cdst {
  int x1, y1, x2, y2;
};
struct fentree {
  vector<int> t;
  fentree() { t.resize(10010); }
  void add(int pos, int x) {
    for (; pos < 10010; pos += (pos & (-pos))) {
      t[pos] += x;
    }
  }
  int sum(int pos) {
    int res = 0;
    for (; pos > 0; pos -= (pos & (-pos))) {
      res += t[pos];
    }
    return res;
  }
  int sum(int l, int r) { return sum(r) - sum(l - 1); }
};
bool comp(cdst a, cdst b) {
  if (a.y2 == b.y2) {
    return a.x1 != a.x2;
  } else
    return a.y2 < b.y2;
}
cdst cs[N];
bool napr[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> cs[i].x1 >> cs[i].y1 >> cs[i].x2 >> cs[i].y2;
    cs[i].x1 += 5001;
    cs[i].y1 += 5001;
    cs[i].x2 += 5001;
    cs[i].y2 += 5001;
    if (cs[i].y1 > cs[i].y2) swap(cs[i].y1, cs[i].y2);
    if (cs[i].x1 > cs[i].x2) swap(cs[i].x1, cs[i].x2);
  }
  sort(cs, cs + n, comp);
  for (int i = 0; i < n; i++) {
    if (cs[i].x1 == cs[i].x2) napr[i] = true;
  }
  fentree ft;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (napr[i]) continue;
    for (int j = 0; j < n; j++) {
      if (napr[j] && cs[i].x1 <= cs[j].x1 && cs[i].x2 >= cs[j].x1 &&
          cs[j].y1 <= cs[i].y1) {
        ft.add(cs[j].x1, 1);
      }
    }
    for (int j = 0; j < n; j++) {
      if (napr[j] && cs[i].x1 <= cs[j].x1 && cs[i].x2 >= cs[j].x1 &&
          cs[j].y1 <= cs[i].y1) {
        ft.add(cs[j].x1, -1);
      } else if (!napr[j] && j > i) {
        long long s = ft.sum(cs[j].x1, cs[j].x2);
        ans += (s * (s - 1)) / 2;
      }
    }
  }
  cout << ans << '\n';
  return 0;
}

```