#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int MAXN = 5e3, MAXM = 1e4 + 1;
int n, cnt1, cnt2, q[MAXN + 5];
long long ans;
struct Node {
  int x1, y1, x2, y2;
  inline friend bool operator<(Node x, Node y) { return x.x1 < y.x1; }
} row[MAXN + 5], col[MAXN + 5];
struct BinaryIndexTree {
  int ind[MAXM + 5];
  inline void clear() { memset(ind, 0, sizeof(ind)); }
  inline void add(int x, int p) {
    for (; x <= MAXM; x += x & -x) ind[x] += p;
  }
  inline int query(int x) {
    int res = 0;
    for (; x; x -= x & -x) res += ind[x];
    return res;
  }
  inline int query(int l, int r) { return query(r) - query(l - 1); }
} bit;
int main() {
  cin >> n;
  for (int x1, y1, x2, y2, i = 1; i <= n; ++i) {
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += MAXN + 1, y1 += MAXN + 1, x2 += MAXN + 1, y2 += MAXN + 1;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (y1 == y2)
      row[++cnt1] = (Node){x1, y1, x2, y2};
    else
      col[++cnt2] = (Node){x1, y1, x2, y2};
  }
  sort(row + 1, row + cnt1 + 1), sort(col + 1, col + cnt2 + 1);
  for (int i = 1; i <= cnt2; ++i) {
    bit.clear();
    int tail = 0;
    for (int j = 1; j <= cnt1; ++j)
      if (row[j].x1 <= col[i].x1 && row[j].x2 >= col[i].x1 &&
          row[j].y1 >= col[i].y1 && row[j].y2 <= col[i].y2) {
        bit.add(row[j].y1, 1);
        q[++tail] = j;
      }
    for (int j = i - 1; j >= 1; --j) {
      for (; row[q[tail]].x1 > col[j].x1 && tail > 0; --tail)
        bit.add(row[q[tail]].y1, -1);
      int l = max(col[i].y1, col[j].y1), r = min(col[i].y2, col[j].y2);
      if (l < r) {
        int res = bit.query(l, r);
        ans += 1ll * res * (res - 1) / 2;
      }
    }
  }
  cout << ans << endl;
}

### 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 mod = 1e9 + 7;
const int MAXN = 5e3, MAXM = 1e4 + 1;
int n, cnt1, cnt2, q[MAXN + 5];
long long ans;
struct Node {
  int x1, y1, x2, y2;
  inline friend bool operator<(Node x, Node y) { return x.x1 < y.x1; }
} row[MAXN + 5], col[MAXN + 5];
struct BinaryIndexTree {
  int ind[MAXM + 5];
  inline void clear() { memset(ind, 0, sizeof(ind)); }
  inline void add(int x, int p) {
    for (; x <= MAXM; x += x & -x) ind[x] += p;
  }
  inline int query(int x) {
    int res = 0;
    for (; x; x -= x & -x) res += ind[x];
    return res;
  }
  inline int query(int l, int r) { return query(r) - query(l - 1); }
} bit;
int main() {
  cin >> n;
  for (int x1, y1, x2, y2, i = 1; i <= n; ++i) {
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += MAXN + 1, y1 += MAXN + 1, x2 += MAXN + 1, y2 += MAXN + 1;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (y1 == y2)
      row[++cnt1] = (Node){x1, y1, x2, y2};
    else
      col[++cnt2] = (Node){x1, y1, x2, y2};
  }
  sort(row + 1, row + cnt1 + 1), sort(col + 1, col + cnt2 + 1);
  for (int i = 1; i <= cnt2; ++i) {
    bit.clear();
    int tail = 0;
    for (int j = 1; j <= cnt1; ++j)
      if (row[j].x1 <= col[i].x1 && row[j].x2 >= col[i].x1 &&
          row[j].y1 >= col[i].y1 && row[j].y2 <= col[i].y2) {
        bit.add(row[j].y1, 1);
        q[++tail] = j;
      }
    for (int j = i - 1; j >= 1; --j) {
      for (; row[q[tail]].x1 > col[j].x1 && tail > 0; --tail)
        bit.add(row[q[tail]].y1, -1);
      int l = max(col[i].y1, col[j].y1), r = min(col[i].y2, col[j].y2);
      if (l < r) {
        int res = bit.query(l, r);
        ans += 1ll * res * (res - 1) / 2;
      }
    }
  }
  cout << ans << endl;
}

```