#include <bits/stdc++.h>
using namespace std;
const int maxn = 5000;
struct Node {
  int x, l, r;
  Node() {}
  Node(int _x, int _l, int _r) { x = _x, l = _l, r = _r; }
  void Output() { printf("x = %d, l = %d, r = %d\n", x, l, r); }
};
vector<Node> sg1, sg2;
vector<pair<int, int>> hi[maxn + 5];
int n, m;
int T = 0;
int tr[maxn * 2 + 5];
int vi[maxn * 2 + 5];
void Add(int x, int d) {
  for (; x <= maxn * 2 + 1; x += x & (-x)) {
    if (vi[x] != T)
      tr[x] = d, vi[x] = T;
    else
      tr[x] += d;
  }
}
int Sum(int x) {
  int r = 0;
  for (; x; x -= x & (-x)) {
    if (vi[x] == T) r += tr[x];
  }
  return r;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x1, y1, x2, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    if (x1 == x2)
      sg1.push_back(Node(x1, min(y1, y2), max(y1, y2)));
    else
      sg2.push_back(Node(y1, min(x1, x2), max(x1, x2)));
  }
  sort(sg1.begin(), sg1.end(),
       [&](const Node &a, const Node &b) { return a.x < b.x; });
  sort(sg2.begin(), sg2.end(),
       [&](const Node &a, const Node &b) { return a.r > b.r; });
  n = sg1.size();
  m = sg2.size();
  for (int i = 0; i < n; ++i) {
    for (int k = 0; k < m; ++k) {
      if (sg1[i].l <= sg2[k].x && sg2[k].x <= sg1[i].r &&
          sg2[k].l <= sg1[i].x && sg1[i].x <= sg2[k].r) {
        hi[i].push_back(pair<int, int>(sg2[k].x, sg2[k].r));
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    int ps = int(hi[i].size()) - 1;
    ++T;
    for (int k = 0; k <= ps; ++k) Add(hi[i][k].first + maxn + 1, 1);
    for (int k = i + 1; k < n; ++k) {
      while (ps >= 0 && hi[i][ps].second < sg1[k].x)
        Add(hi[i][ps].first + maxn + 1, -1), --ps;
      if (ps + 1 <= 1) break;
      if (sg1[i].x == sg1[k].x) continue;
      int ct = Sum(sg1[k].r + maxn + 1) - Sum(sg1[k].l - 1 + maxn + 1);
      ans += ((ct * (ct - 1)) >> 1);
    }
  }
  printf("%lld\n", ans);
}

### 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 maxn = 5000;
struct Node {
  int x, l, r;
  Node() {}
  Node(int _x, int _l, int _r) { x = _x, l = _l, r = _r; }
  void Output() { printf("x = %d, l = %d, r = %d\n", x, l, r); }
};
vector<Node> sg1, sg2;
vector<pair<int, int>> hi[maxn + 5];
int n, m;
int T = 0;
int tr[maxn * 2 + 5];
int vi[maxn * 2 + 5];
void Add(int x, int d) {
  for (; x <= maxn * 2 + 1; x += x & (-x)) {
    if (vi[x] != T)
      tr[x] = d, vi[x] = T;
    else
      tr[x] += d;
  }
}
int Sum(int x) {
  int r = 0;
  for (; x; x -= x & (-x)) {
    if (vi[x] == T) r += tr[x];
  }
  return r;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x1, y1, x2, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    if (x1 == x2)
      sg1.push_back(Node(x1, min(y1, y2), max(y1, y2)));
    else
      sg2.push_back(Node(y1, min(x1, x2), max(x1, x2)));
  }
  sort(sg1.begin(), sg1.end(),
       [&](const Node &a, const Node &b) { return a.x < b.x; });
  sort(sg2.begin(), sg2.end(),
       [&](const Node &a, const Node &b) { return a.r > b.r; });
  n = sg1.size();
  m = sg2.size();
  for (int i = 0; i < n; ++i) {
    for (int k = 0; k < m; ++k) {
      if (sg1[i].l <= sg2[k].x && sg2[k].x <= sg1[i].r &&
          sg2[k].l <= sg1[i].x && sg1[i].x <= sg2[k].r) {
        hi[i].push_back(pair<int, int>(sg2[k].x, sg2[k].r));
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    int ps = int(hi[i].size()) - 1;
    ++T;
    for (int k = 0; k <= ps; ++k) Add(hi[i][k].first + maxn + 1, 1);
    for (int k = i + 1; k < n; ++k) {
      while (ps >= 0 && hi[i][ps].second < sg1[k].x)
        Add(hi[i][ps].first + maxn + 1, -1), --ps;
      if (ps + 1 <= 1) break;
      if (sg1[i].x == sg1[k].x) continue;
      int ct = Sum(sg1[k].r + maxn + 1) - Sum(sg1[k].l - 1 + maxn + 1);
      ans += ((ct * (ct - 1)) >> 1);
    }
  }
  printf("%lld\n", ans);
}

```