#include <bits/stdc++.h>
using namespace std;
const int N = 5e3 + 5;
int n, a, b, x, y, cnt1 = 0, cnt2 = 0;
bitset<N> f[N];
long long Ans = 0;
struct data {
  int a, b, x, y;
} heng[N], shu[N];
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return f * x;
}
int jiao(data k1, data k2) {
  int s1 = 0, s2 = 0;
  if (k1.a <= k2.x && k2.x <= k1.x && k1.a <= k2.a && k2.x <= k1.x) s1 = 1;
  if (k2.b <= k1.b && k1.b <= k2.y && k2.b <= k1.y && k1.y <= k2.y) s2 = 1;
  return s1 & s2;
}
int main() {
  n = read();
  for (register int i = 1; i <= n; i++) {
    a = read(), b = read(), x = read(), y = read();
    if (a > x) swap(a, x);
    if (b > y) swap(b, y);
    if (a == x) shu[++cnt1] = (data){a, b, x, y};
    if (b == y) heng[++cnt2] = (data){a, b, x, y};
  }
  if (cnt1 > cnt2) {
    for (register int i = 1; i <= cnt2; i++)
      for (register int j = 1; j <= cnt1; j++)
        if (jiao(heng[i], shu[j])) f[i][j] = 1;
    for (register int i = 1; i <= cnt2 - 1; i++)
      for (register int j = i + 1; j <= cnt2; j++) {
        int tmp = (f[i] & f[j]).count();
        if (tmp >= 2) Ans += 1LL * (tmp - 1) * tmp / 2;
      }
  } else {
    for (register int i = 1; i <= cnt1; i++)
      for (register int j = 1; j <= cnt2; j++)
        if (jiao(heng[j], shu[i])) f[i][j] = 1;
    for (register int i = 1; i <= cnt1 - 1; i++)
      for (register int j = i + 1; j <= cnt1; j++) {
        int tmp = (f[i] & f[j]).count();
        if (tmp >= 2) Ans += 1LL * (tmp - 1) * tmp / 2;
      }
  }
  printf("%lld\n", Ans);
  return 0;
}

### 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;
const int N = 5e3 + 5;
int n, a, b, x, y, cnt1 = 0, cnt2 = 0;
bitset<N> f[N];
long long Ans = 0;
struct data {
  int a, b, x, y;
} heng[N], shu[N];
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return f * x;
}
int jiao(data k1, data k2) {
  int s1 = 0, s2 = 0;
  if (k1.a <= k2.x && k2.x <= k1.x && k1.a <= k2.a && k2.x <= k1.x) s1 = 1;
  if (k2.b <= k1.b && k1.b <= k2.y && k2.b <= k1.y && k1.y <= k2.y) s2 = 1;
  return s1 & s2;
}
int main() {
  n = read();
  for (register int i = 1; i <= n; i++) {
    a = read(), b = read(), x = read(), y = read();
    if (a > x) swap(a, x);
    if (b > y) swap(b, y);
    if (a == x) shu[++cnt1] = (data){a, b, x, y};
    if (b == y) heng[++cnt2] = (data){a, b, x, y};
  }
  if (cnt1 > cnt2) {
    for (register int i = 1; i <= cnt2; i++)
      for (register int j = 1; j <= cnt1; j++)
        if (jiao(heng[i], shu[j])) f[i][j] = 1;
    for (register int i = 1; i <= cnt2 - 1; i++)
      for (register int j = i + 1; j <= cnt2; j++) {
        int tmp = (f[i] & f[j]).count();
        if (tmp >= 2) Ans += 1LL * (tmp - 1) * tmp / 2;
      }
  } else {
    for (register int i = 1; i <= cnt1; i++)
      for (register int j = 1; j <= cnt2; j++)
        if (jiao(heng[j], shu[i])) f[i][j] = 1;
    for (register int i = 1; i <= cnt1 - 1; i++)
      for (register int j = i + 1; j <= cnt1; j++) {
        int tmp = (f[i] & f[j]).count();
        if (tmp >= 2) Ans += 1LL * (tmp - 1) * tmp / 2;
      }
  }
  printf("%lld\n", Ans);
  return 0;
}

```