#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 233;
long long ans = 0;
int ch[maxn], N = 10010;
void add(int x, int v) {
  for (int i = x; i <= N; i += i & (-i)) ch[i] += v;
}
int query(int x) {
  int sum = 0;
  for (int i = x; i; i -= i & (-i)) sum += ch[i];
  return sum;
}
struct H {
  int l, r, y;
  H() {}
  H(int l, int r, int y) : l(l), r(r), y(y) {}
} h[maxn];
struct S {
  int u, d, x;
  S() {}
  S(int d, int u, int x) : u(u), d(d), x(x) {}
} s[maxn];
bool cmp1(const H& a, const H& b) { return a.r < b.r; }
bool cmp2(const S& a, const S& b) { return a.x < b.x; }
int a = 0, b = 0, cur, pre;
void init(int x) {
  memset(ch, 0, sizeof(ch));
  for (int i = 1; i <= a; i++) {
    if (h[i].l <= x) add(h[i].y, 1);
  }
  pre = x;
  cur = 1;
}
void remove(int x) {
  while (cur <= a && h[cur].r < x) {
    if (h[cur].l <= pre) add(h[cur].y, -1);
    cur++;
  }
}
void cal(int l, int r) {
  if (l >= r) return;
  int len = query(r) - query(l - 1);
  ans += len * (len - 1) / 2;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x1, x2, y1, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += 5001;
    y1 += 5001;
    x2 += 5001;
    y2 += 5001;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2)
      s[++b] = S(y1, y2, x1);
    else
      h[++a] = H(x1, x2, y1);
  }
  sort(h + 1, h + 1 + a, cmp1);
  sort(s + 1, s + 1 + b, cmp2);
  for (int i = 1; i <= b; i++) {
    init(s[i].x);
    for (int j = i + 1; j <= b; j++) {
      remove(s[j].x);
      int U = min(s[i].u, s[j].u);
      int D = max(s[i].d, s[j].d);
      cal(D, U);
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 = 1e4 + 233;
long long ans = 0;
int ch[maxn], N = 10010;
void add(int x, int v) {
  for (int i = x; i <= N; i += i & (-i)) ch[i] += v;
}
int query(int x) {
  int sum = 0;
  for (int i = x; i; i -= i & (-i)) sum += ch[i];
  return sum;
}
struct H {
  int l, r, y;
  H() {}
  H(int l, int r, int y) : l(l), r(r), y(y) {}
} h[maxn];
struct S {
  int u, d, x;
  S() {}
  S(int d, int u, int x) : u(u), d(d), x(x) {}
} s[maxn];
bool cmp1(const H& a, const H& b) { return a.r < b.r; }
bool cmp2(const S& a, const S& b) { return a.x < b.x; }
int a = 0, b = 0, cur, pre;
void init(int x) {
  memset(ch, 0, sizeof(ch));
  for (int i = 1; i <= a; i++) {
    if (h[i].l <= x) add(h[i].y, 1);
  }
  pre = x;
  cur = 1;
}
void remove(int x) {
  while (cur <= a && h[cur].r < x) {
    if (h[cur].l <= pre) add(h[cur].y, -1);
    cur++;
  }
}
void cal(int l, int r) {
  if (l >= r) return;
  int len = query(r) - query(l - 1);
  ans += len * (len - 1) / 2;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x1, x2, y1, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += 5001;
    y1 += 5001;
    x2 += 5001;
    y2 += 5001;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2)
      s[++b] = S(y1, y2, x1);
    else
      h[++a] = H(x1, x2, y1);
  }
  sort(h + 1, h + 1 + a, cmp1);
  sort(s + 1, s + 1 + b, cmp2);
  for (int i = 1; i <= b; i++) {
    init(s[i].x);
    for (int j = i + 1; j <= b; j++) {
      remove(s[j].x);
      int U = min(s[i].u, s[j].u);
      int D = max(s[i].d, s[j].d);
      cal(D, U);
    }
  }
  cout << ans << endl;
  return 0;
}

```