#include <bits/stdc++.h>
using namespace std;
const int maxn = 5000 + 10;
struct HLine {
  int xa, xb, y;
  bool operator<(const HLine& rhs) const { return y < rhs.y; }
};
struct Node {
  int y, x, flag;
  bool operator<(const Node& rhs) const { return y < rhs.y; }
};
int cnt;
vector<HLine> h;
vector<Node> a;
int x[2 * maxn];
int c[2 * maxn];
int vis[2 * maxn];
void add(int pos, int val) {
  for (int i = pos; i <= cnt; i += i & (-i)) c[i] += val;
}
int sum(int pos) {
  int res = 0;
  for (int i = pos; i; i -= i & (-i)) res += c[i];
  return res;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int xa, ya, xb, yb;
    scanf("%d%d%d%d", &xa, &ya, &xb, &yb);
    x[++cnt] = xa;
    x[++cnt] = xb;
    if (xa == xb) {
      a.push_back((Node){min(ya, yb), xa, 1});
      a.push_back((Node){max(ya, yb), xa, -1});
    } else
      h.push_back((HLine){min(xa, xb), max(xa, xb), ya});
  }
  sort(a.begin(), a.end());
  sort(h.begin(), h.end());
  sort(x + 1, x + cnt + 1);
  cnt = unique(x + 1, x + cnt + 1) - (x + 1);
  for (int i = 0; i < h.size(); i++) {
    h[i].xa = lower_bound(x + 1, x + cnt + 1, h[i].xa) - x;
    h[i].xb = lower_bound(x + 1, x + cnt + 1, h[i].xb) - x;
  }
  for (int i = 0; i < a.size(); i++)
    a[i].x = lower_bound(x + 1, x + cnt + 1, a[i].x) - x;
  long long ans = 0;
  for (int i = 0; i < h.size(); i++) {
    int pos = -1;
    while (pos + 1 < a.size() && a[pos + 1].y <= h[i].y) {
      pos++;
      if (a[pos].flag < 0)
        add(a[pos].x, -1);
      else
        add(a[pos].x, 1);
    }
    for (int j = i + 1; j < h.size(); j++) {
      while (pos + 1 < a.size() && a[pos + 1].y < h[j].y) {
        pos++;
        if (a[pos].flag < 0) {
          if (vis[a[pos].x])
            vis[a[pos].x]--;
          else
            add(a[pos].x, -1);
        } else
          vis[a[pos].x]++;
      }
      int L = max(h[i].xa, h[j].xa);
      int R = min(h[i].xb, h[j].xb);
      if (R < L) continue;
      int num = sum(R) - sum(L - 1);
      ans += num * (num - 1) / 2;
    }
    while (pos + 1 < a.size()) {
      pos++;
      if (a[pos].flag < 0) {
        if (vis[a[pos].x])
          vis[a[pos].x]--;
        else
          add(a[pos].x, -1);
      } else
        vis[a[pos].x]++;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Generate 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 + 10;
struct HLine {
  int xa, xb, y;
  bool operator<(const HLine& rhs) const { return y < rhs.y; }
};
struct Node {
  int y, x, flag;
  bool operator<(const Node& rhs) const { return y < rhs.y; }
};
int cnt;
vector<HLine> h;
vector<Node> a;
int x[2 * maxn];
int c[2 * maxn];
int vis[2 * maxn];
void add(int pos, int val) {
  for (int i = pos; i <= cnt; i += i & (-i)) c[i] += val;
}
int sum(int pos) {
  int res = 0;
  for (int i = pos; i; i -= i & (-i)) res += c[i];
  return res;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int xa, ya, xb, yb;
    scanf("%d%d%d%d", &xa, &ya, &xb, &yb);
    x[++cnt] = xa;
    x[++cnt] = xb;
    if (xa == xb) {
      a.push_back((Node){min(ya, yb), xa, 1});
      a.push_back((Node){max(ya, yb), xa, -1});
    } else
      h.push_back((HLine){min(xa, xb), max(xa, xb), ya});
  }
  sort(a.begin(), a.end());
  sort(h.begin(), h.end());
  sort(x + 1, x + cnt + 1);
  cnt = unique(x + 1, x + cnt + 1) - (x + 1);
  for (int i = 0; i < h.size(); i++) {
    h[i].xa = lower_bound(x + 1, x + cnt + 1, h[i].xa) - x;
    h[i].xb = lower_bound(x + 1, x + cnt + 1, h[i].xb) - x;
  }
  for (int i = 0; i < a.size(); i++)
    a[i].x = lower_bound(x + 1, x + cnt + 1, a[i].x) - x;
  long long ans = 0;
  for (int i = 0; i < h.size(); i++) {
    int pos = -1;
    while (pos + 1 < a.size() && a[pos + 1].y <= h[i].y) {
      pos++;
      if (a[pos].flag < 0)
        add(a[pos].x, -1);
      else
        add(a[pos].x, 1);
    }
    for (int j = i + 1; j < h.size(); j++) {
      while (pos + 1 < a.size() && a[pos + 1].y < h[j].y) {
        pos++;
        if (a[pos].flag < 0) {
          if (vis[a[pos].x])
            vis[a[pos].x]--;
          else
            add(a[pos].x, -1);
        } else
          vis[a[pos].x]++;
      }
      int L = max(h[i].xa, h[j].xa);
      int R = min(h[i].xb, h[j].xb);
      if (R < L) continue;
      int num = sum(R) - sum(L - 1);
      ans += num * (num - 1) / 2;
    }
    while (pos + 1 < a.size()) {
      pos++;
      if (a[pos].flag < 0) {
        if (vis[a[pos].x])
          vis[a[pos].x]--;
        else
          add(a[pos].x, -1);
      } else
        vis[a[pos].x]++;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```