#include <bits/stdc++.h>
using namespace std;
const int N = 50005;
int n;
int x[N][2], y[N][2];
int X[N], Y[N];
struct seg1 {
  int x, y1, y2;
  bool operator<(const seg1 &A) const { return x < A.x; }
} s1[N];
struct seg2 {
  int y, x1, x2;
} s2[N];
vector<int> v1[N], v2[N], v3[N];
int c[N];
int lowbit(int x) { return x & (-x); }
void add(int x, int v) {
  for (int i = x; i < N; i += lowbit(i)) c[i] += v;
}
int query(int p) {
  int ans = 0;
  for (int i = p; i; i -= lowbit(i)) ans += c[i];
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int D1 = 0, D2 = 0;
  for (int i = 1; i <= n; i++) {
    cin >> x[i][0] >> y[i][0] >> x[i][1] >> y[i][1];
    X[++D1] = x[i][0];
    X[++D1] = x[i][1];
    Y[++D2] = y[i][0];
    Y[++D2] = y[i][1];
  }
  sort(X + 1, X + D1 + 1);
  sort(Y + 1, Y + D2 + 1);
  D1 = unique(X + 1, X + D1 + 1) - X - 1;
  D2 = unique(Y + 1, Y + D2 + 1) - Y - 1;
  int t1 = 0, t2 = 0;
  for (int i = 1; i <= n; i++) {
    x[i][0] = lower_bound(X + 1, X + D1 + 1, x[i][0]) - X;
    x[i][1] = lower_bound(X + 1, X + D1 + 1, x[i][1]) - X;
    y[i][0] = lower_bound(Y + 1, Y + D2 + 1, y[i][0]) - Y;
    y[i][1] = lower_bound(Y + 1, Y + D2 + 1, y[i][1]) - Y;
    if (x[i][0] > x[i][1]) swap(x[i][0], x[i][1]);
    if (y[i][0] > y[i][1]) swap(y[i][0], y[i][1]);
    if (x[i][0] == x[i][1]) s1[++t1] = seg1{x[i][0], y[i][0], y[i][1]};
    if (y[i][0] == y[i][1]) s2[++t2] = seg2{y[i][0], x[i][0], x[i][1]};
    if (y[i][0] == y[i][1]) v1[x[i][0]].push_back(t2);
  }
  sort(s1 + 1, s1 + t1 + 1);
  for (int i = 1; i <= t1; i++) v3[s1[i].x].push_back(i);
  long long res = 0;
  int bound = D1;
  for (int x = 1; x <= bound; x++) {
    for (auto it : v2[x]) {
      add(s2[it].y, -1);
    }
    for (auto it : v1[x]) {
      add(s2[it].y, 1);
      v2[s2[it].x2].push_back(it);
    }
    for (auto i : v3[x]) {
      for (int j = x + 1; j <= bound; j++) {
        for (auto k : v3[j]) {
          int R = min(s1[i].y2, s1[k].y2), L = max(s1[i].y1, s1[k].y1);
          if (L > R) continue;
          int tmp = query(R) - query(L - 1);
          res += 1ll * tmp * (tmp - 1) / 2;
        }
        for (auto it : v2[j]) add(s2[it].y, -1);
      }
      for (int j = x + 1; j <= bound; j++) {
        for (auto it : v2[j]) add(s2[it].y, 1);
      }
    }
  }
  cout << res;
  return 0;
}

### Prompt

Please create a solution in Cpp 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 N = 50005;
int n;
int x[N][2], y[N][2];
int X[N], Y[N];
struct seg1 {
  int x, y1, y2;
  bool operator<(const seg1 &A) const { return x < A.x; }
} s1[N];
struct seg2 {
  int y, x1, x2;
} s2[N];
vector<int> v1[N], v2[N], v3[N];
int c[N];
int lowbit(int x) { return x & (-x); }
void add(int x, int v) {
  for (int i = x; i < N; i += lowbit(i)) c[i] += v;
}
int query(int p) {
  int ans = 0;
  for (int i = p; i; i -= lowbit(i)) ans += c[i];
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int D1 = 0, D2 = 0;
  for (int i = 1; i <= n; i++) {
    cin >> x[i][0] >> y[i][0] >> x[i][1] >> y[i][1];
    X[++D1] = x[i][0];
    X[++D1] = x[i][1];
    Y[++D2] = y[i][0];
    Y[++D2] = y[i][1];
  }
  sort(X + 1, X + D1 + 1);
  sort(Y + 1, Y + D2 + 1);
  D1 = unique(X + 1, X + D1 + 1) - X - 1;
  D2 = unique(Y + 1, Y + D2 + 1) - Y - 1;
  int t1 = 0, t2 = 0;
  for (int i = 1; i <= n; i++) {
    x[i][0] = lower_bound(X + 1, X + D1 + 1, x[i][0]) - X;
    x[i][1] = lower_bound(X + 1, X + D1 + 1, x[i][1]) - X;
    y[i][0] = lower_bound(Y + 1, Y + D2 + 1, y[i][0]) - Y;
    y[i][1] = lower_bound(Y + 1, Y + D2 + 1, y[i][1]) - Y;
    if (x[i][0] > x[i][1]) swap(x[i][0], x[i][1]);
    if (y[i][0] > y[i][1]) swap(y[i][0], y[i][1]);
    if (x[i][0] == x[i][1]) s1[++t1] = seg1{x[i][0], y[i][0], y[i][1]};
    if (y[i][0] == y[i][1]) s2[++t2] = seg2{y[i][0], x[i][0], x[i][1]};
    if (y[i][0] == y[i][1]) v1[x[i][0]].push_back(t2);
  }
  sort(s1 + 1, s1 + t1 + 1);
  for (int i = 1; i <= t1; i++) v3[s1[i].x].push_back(i);
  long long res = 0;
  int bound = D1;
  for (int x = 1; x <= bound; x++) {
    for (auto it : v2[x]) {
      add(s2[it].y, -1);
    }
    for (auto it : v1[x]) {
      add(s2[it].y, 1);
      v2[s2[it].x2].push_back(it);
    }
    for (auto i : v3[x]) {
      for (int j = x + 1; j <= bound; j++) {
        for (auto k : v3[j]) {
          int R = min(s1[i].y2, s1[k].y2), L = max(s1[i].y1, s1[k].y1);
          if (L > R) continue;
          int tmp = query(R) - query(L - 1);
          res += 1ll * tmp * (tmp - 1) / 2;
        }
        for (auto it : v2[j]) add(s2[it].y, -1);
      }
      for (int j = x + 1; j <= bound; j++) {
        for (auto it : v2[j]) add(s2[it].y, 1);
      }
    }
  }
  cout << res;
  return 0;
}

```