#include <bits/stdc++.h>
using namespace std;
const int maxn = 5010;
struct Line {
  int x1, y1, x2, y2;
} H[maxn], V[maxn], VT[maxn];
int hn, vn;
int f[maxn * 2];
long long ans;
int n;
bool cmpx(const Line &a, const Line &b) { return a.x1 < b.x1; }
bool cmpy(const Line &a, const Line &b) { return a.y1 < b.y1; }
bool cmpy2(const Line &a, const Line &b) { return a.y2 < b.y2; }
int lowbit(int x) { return x & -x; }
void Add(int x, int v) {
  while (x <= 10001) f[x] += v, x += lowbit(x);
}
int Query(int x) {
  int res = 0;
  while (x) res += f[x], x -= lowbit(x);
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int tx1, ty1, tx2, ty2;
    scanf("%d%d%d%d", &tx1, &ty1, &tx2, &ty2);
    if (tx1 == tx2) {
      ++vn;
      if (ty1 > ty2) swap(ty1, ty2);
      V[vn].x1 = tx1;
      V[vn].y1 = ty1;
      V[vn].x2 = tx2;
      V[vn].y2 = ty2;
    } else {
      ++hn;
      if (tx1 > tx2) swap(tx1, tx2);
      H[hn].x1 = tx1;
      H[hn].y1 = ty1;
      H[hn].x2 = tx2;
      H[hn].y2 = ty2;
    }
  }
  sort(H + 1, H + 1 + hn, cmpy);
  for (int i = 1; i <= hn; i++) {
    int m = 0;
    for (int j = 1; j <= vn; j++)
      if (V[j].y1 <= H[i].y1 && V[j].y2 >= H[i].y1) VT[++m] = V[j];
    sort(VT + 1, VT + 1 + m, cmpy2);
    for (int j = 1; j <= m; j++) Add(VT[j].x1 + 5001, 1);
    int k = 1;
    for (int j = i + 1; j <= hn; j++) {
      while (k <= m && VT[k].y2 < H[j].y1) Add(VT[k].x1 + 5001, -1), k++;
      int xl = max(H[i].x1, H[j].x1), xr = min(H[i].x2, H[j].x2);
      if (xr < xl) continue;
      int res = Query(xr + 5001) - Query(xl + 5000);
      ans += (long long)res * (res - 1) / 2;
    }
    while (k <= m) Add(VT[k].x1 + 5001, -1), k++;
  }
  printf("%lld\n", ans);
}

### Prompt

Please create a solution in Cpp 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 = 5010;
struct Line {
  int x1, y1, x2, y2;
} H[maxn], V[maxn], VT[maxn];
int hn, vn;
int f[maxn * 2];
long long ans;
int n;
bool cmpx(const Line &a, const Line &b) { return a.x1 < b.x1; }
bool cmpy(const Line &a, const Line &b) { return a.y1 < b.y1; }
bool cmpy2(const Line &a, const Line &b) { return a.y2 < b.y2; }
int lowbit(int x) { return x & -x; }
void Add(int x, int v) {
  while (x <= 10001) f[x] += v, x += lowbit(x);
}
int Query(int x) {
  int res = 0;
  while (x) res += f[x], x -= lowbit(x);
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int tx1, ty1, tx2, ty2;
    scanf("%d%d%d%d", &tx1, &ty1, &tx2, &ty2);
    if (tx1 == tx2) {
      ++vn;
      if (ty1 > ty2) swap(ty1, ty2);
      V[vn].x1 = tx1;
      V[vn].y1 = ty1;
      V[vn].x2 = tx2;
      V[vn].y2 = ty2;
    } else {
      ++hn;
      if (tx1 > tx2) swap(tx1, tx2);
      H[hn].x1 = tx1;
      H[hn].y1 = ty1;
      H[hn].x2 = tx2;
      H[hn].y2 = ty2;
    }
  }
  sort(H + 1, H + 1 + hn, cmpy);
  for (int i = 1; i <= hn; i++) {
    int m = 0;
    for (int j = 1; j <= vn; j++)
      if (V[j].y1 <= H[i].y1 && V[j].y2 >= H[i].y1) VT[++m] = V[j];
    sort(VT + 1, VT + 1 + m, cmpy2);
    for (int j = 1; j <= m; j++) Add(VT[j].x1 + 5001, 1);
    int k = 1;
    for (int j = i + 1; j <= hn; j++) {
      while (k <= m && VT[k].y2 < H[j].y1) Add(VT[k].x1 + 5001, -1), k++;
      int xl = max(H[i].x1, H[j].x1), xr = min(H[i].x2, H[j].x2);
      if (xr < xl) continue;
      int res = Query(xr + 5001) - Query(xl + 5000);
      ans += (long long)res * (res - 1) / 2;
    }
    while (k <= m) Add(VT[k].x1 + 5001, -1), k++;
  }
  printf("%lld\n", ans);
}

```