#include <bits/stdc++.h>
using namespace std;
const int N = 100001;
int seg[N], hor[N], x[N][2], y[N][2], p[N];
void update(int r, int add) {
  while (r < N) {
    seg[r] += add;
    r += r & (-r);
  }
}
int query(int r) {
  int ans = 0;
  while (r < N && r > 0) {
    ans += seg[r];
    r -= r & (-r);
  }
  return ans;
}
bool cmp(int i, int j) {
  if (x[i][0] != x[j][0]) {
    return x[i][0] < x[j][0];
  }
  if (hor[i] + hor[j] <= 1) {
    return hor[i];
  }
  return x[i][1] < x[j][1];
}
int n;
long long ans = 0;
void calc(int idx) {
  memset(seg, 0, sizeof(seg));
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      q;
  for (int i = 1; i < idx; i++) {
    if (hor[p[i]]) {
      update(y[p[i]][0], 1);
      q.push({x[p[i]][1] + 1, i});
    }
  }
  for (int i = idx + 1; i <= n; i++) {
    while (q.size() && q.top().first <= x[p[i]][0]) {
      update(y[p[q.top().second]][0], -1);
      q.pop();
    }
    int cnt = query(min(y[p[idx]][1], y[p[i]][1])) -
              query(max(y[p[idx]][0], y[p[i]][0]) - 1);
    if (cnt > 0) {
      ans += 1LL * cnt * (cnt - 1) / 2;
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 2; j++) {
      scanf("%d %d", &x[i][j], &y[i][j]);
      x[i][j] += 5001, y[i][j] += 5001;
    }
    if (x[i][0] == x[i][1]) {
      hor[i] = 0;
    } else {
      hor[i] = 1;
    }
    if (x[i][0] > x[i][1]) {
      swap(x[i][0], x[i][1]);
      swap(y[i][0], y[i][1]);
    } else if (y[i][0] > y[i][1]) {
      swap(x[i][0], x[i][1]);
      swap(y[i][0], y[i][1]);
    }
    p[i] = i;
  }
  sort(p + 1, p + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    if (hor[p[i]] == 0) {
      calc(i);
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your task is to create 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 N = 100001;
int seg[N], hor[N], x[N][2], y[N][2], p[N];
void update(int r, int add) {
  while (r < N) {
    seg[r] += add;
    r += r & (-r);
  }
}
int query(int r) {
  int ans = 0;
  while (r < N && r > 0) {
    ans += seg[r];
    r -= r & (-r);
  }
  return ans;
}
bool cmp(int i, int j) {
  if (x[i][0] != x[j][0]) {
    return x[i][0] < x[j][0];
  }
  if (hor[i] + hor[j] <= 1) {
    return hor[i];
  }
  return x[i][1] < x[j][1];
}
int n;
long long ans = 0;
void calc(int idx) {
  memset(seg, 0, sizeof(seg));
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      q;
  for (int i = 1; i < idx; i++) {
    if (hor[p[i]]) {
      update(y[p[i]][0], 1);
      q.push({x[p[i]][1] + 1, i});
    }
  }
  for (int i = idx + 1; i <= n; i++) {
    while (q.size() && q.top().first <= x[p[i]][0]) {
      update(y[p[q.top().second]][0], -1);
      q.pop();
    }
    int cnt = query(min(y[p[idx]][1], y[p[i]][1])) -
              query(max(y[p[idx]][0], y[p[i]][0]) - 1);
    if (cnt > 0) {
      ans += 1LL * cnt * (cnt - 1) / 2;
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 2; j++) {
      scanf("%d %d", &x[i][j], &y[i][j]);
      x[i][j] += 5001, y[i][j] += 5001;
    }
    if (x[i][0] == x[i][1]) {
      hor[i] = 0;
    } else {
      hor[i] = 1;
    }
    if (x[i][0] > x[i][1]) {
      swap(x[i][0], x[i][1]);
      swap(y[i][0], y[i][1]);
    } else if (y[i][0] > y[i][1]) {
      swap(x[i][0], x[i][1]);
      swap(y[i][0], y[i][1]);
    }
    p[i] = i;
  }
  sort(p + 1, p + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    if (hor[p[i]] == 0) {
      calc(i);
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```