#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
const int mod = 1e9 + 7;
const int maxn = 5e3 + 5;
struct Horizontal {
  int y, x1, x2;
  Horizontal() {}
  Horizontal(int p, int x1, int x2) : y(p), x1(x1), x2(x2) {}
  bool operator<(const Horizontal &b) const { return y < b.y; }
};
struct Vertical {
  int x, y1, y2;
  Vertical() {}
  Vertical(int p, int y1, int y2) : x(p), y1(y1), y2(y2) {}
  bool operator<(const Vertical &b) const { return y2 < b.y2; }
};
int ch, cv;
Horizontal h[maxn];
int qu[maxn], b, f;
Vertical v[maxn];
int temp[maxn * 2], cnt = 1, m;
ll tree[maxn * 2];
void add(int x, int in) {
  while (x <= m) {
    tree[x] += in;
    x += x & -x;
  }
}
ll query(int x) {
  ll re = 1;
  while (x > 0) {
    re += tree[x];
    x -= x & -x;
  }
  return re;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    int x1, x2, y1, y2;
    scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
    if (x1 == x2) {
      v[cv++] = Vertical(x1, min(y1, y2), max(y1, y2));
    } else if (y1 == y2) {
      h[ch++] = Horizontal(y1, min(x1, x2), max(x1, x2));
    }
    temp[cnt++] = x1, temp[cnt++] = x2;
  }
  sort(temp + 1, temp + cnt);
  m = unique(temp + 1, temp + cnt) - temp;
  for (int i = 0; i < ch; ++i) {
    h[i].x1 = lower_bound(temp + 1, temp + m, h[i].x1) - temp;
    h[i].x2 = lower_bound(temp + 1, temp + m, h[i].x2) - temp;
  }
  for (int i = 0; i < cv; ++i) {
    v[i].x = lower_bound(temp + 1, temp + m, v[i].x) - temp;
  }
  ll ans = 0;
  sort(h, h + ch);
  sort(v, v + cv);
  for (int i = 0; i < ch; ++i) {
    memset(tree, 0, sizeof(tree));
    b = 0;
    f = 1;
    for (int c = 0; c < cv; ++c) {
      if (h[i].y <= v[c].y2 && h[i].y >= v[c].y1) {
        add(v[c].x, 1);
        qu[++b] = c;
      }
    }
    for (int j = i + 1; j < ch; ++j) {
      if (h[i].y == h[j].y) continue;
      while (f <= b && v[qu[f]].y2 < h[j].y) {
        add(v[qu[f]].x, -1);
        ++f;
      }
      int l, r;
      l = max(h[i].x1, h[j].x1);
      r = min(h[i].x2, h[j].x2);
      if (l >= r) continue;
      ll add = query(r) - query(l - 1);
      ans += add * (add - 1) / 2;
    }
  }
  printf("%lld\n", ans);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
const int mod = 1e9 + 7;
const int maxn = 5e3 + 5;
struct Horizontal {
  int y, x1, x2;
  Horizontal() {}
  Horizontal(int p, int x1, int x2) : y(p), x1(x1), x2(x2) {}
  bool operator<(const Horizontal &b) const { return y < b.y; }
};
struct Vertical {
  int x, y1, y2;
  Vertical() {}
  Vertical(int p, int y1, int y2) : x(p), y1(y1), y2(y2) {}
  bool operator<(const Vertical &b) const { return y2 < b.y2; }
};
int ch, cv;
Horizontal h[maxn];
int qu[maxn], b, f;
Vertical v[maxn];
int temp[maxn * 2], cnt = 1, m;
ll tree[maxn * 2];
void add(int x, int in) {
  while (x <= m) {
    tree[x] += in;
    x += x & -x;
  }
}
ll query(int x) {
  ll re = 1;
  while (x > 0) {
    re += tree[x];
    x -= x & -x;
  }
  return re;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    int x1, x2, y1, y2;
    scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
    if (x1 == x2) {
      v[cv++] = Vertical(x1, min(y1, y2), max(y1, y2));
    } else if (y1 == y2) {
      h[ch++] = Horizontal(y1, min(x1, x2), max(x1, x2));
    }
    temp[cnt++] = x1, temp[cnt++] = x2;
  }
  sort(temp + 1, temp + cnt);
  m = unique(temp + 1, temp + cnt) - temp;
  for (int i = 0; i < ch; ++i) {
    h[i].x1 = lower_bound(temp + 1, temp + m, h[i].x1) - temp;
    h[i].x2 = lower_bound(temp + 1, temp + m, h[i].x2) - temp;
  }
  for (int i = 0; i < cv; ++i) {
    v[i].x = lower_bound(temp + 1, temp + m, v[i].x) - temp;
  }
  ll ans = 0;
  sort(h, h + ch);
  sort(v, v + cv);
  for (int i = 0; i < ch; ++i) {
    memset(tree, 0, sizeof(tree));
    b = 0;
    f = 1;
    for (int c = 0; c < cv; ++c) {
      if (h[i].y <= v[c].y2 && h[i].y >= v[c].y1) {
        add(v[c].x, 1);
        qu[++b] = c;
      }
    }
    for (int j = i + 1; j < ch; ++j) {
      if (h[i].y == h[j].y) continue;
      while (f <= b && v[qu[f]].y2 < h[j].y) {
        add(v[qu[f]].x, -1);
        ++f;
      }
      int l, r;
      l = max(h[i].x1, h[j].x1);
      r = min(h[i].x2, h[j].x2);
      if (l >= r) continue;
      ll add = query(r) - query(l - 1);
      ans += add * (add - 1) / 2;
    }
  }
  printf("%lld\n", ans);
}

```