#include <bits/stdc++.h>
using namespace std;
inline int read() {
  char ch = getchar();
  int f = 1, x = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f * x;
}
const int N = 10001;
int n, t[N + 1];
vector<int> cur[N + 1];
vector<pair<int, int> > row[N + 1], col[N + 1];
inline void init() {
  memset(t, 0, sizeof(t));
  for (int i = 1; i <= N; i++) cur[i].clear();
}
inline void add(int x, int y) {
  for (; x <= N; x += x & -x) t[x] += y;
}
inline int query(int x) {
  int ans = 0;
  for (; x; x -= x & -x) ans += t[x];
  return ans;
}
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int x1, y1, x2, y2;
    x1 = read() + 5001;
    y1 = read() + 5001;
    x2 = read() + 5001;
    y2 = read() + 5001;
    if (x1 > x2) x1 ^= x2 ^= x1 ^= x2;
    if (y1 > y2) y1 ^= y2 ^= y1 ^= y2;
    if (y1 == y2)
      row[y1].emplace_back(make_pair(x1, x2));
    else
      col[x1].emplace_back(make_pair(y1, y2));
  }
  long long ans = 0;
  for (int y = 1; y <= N; y++) {
    for (auto i : row[y]) {
      init();
      int lx = i.first, rx = i.second;
      for (int x = lx; x <= rx; x++) {
        for (auto j : col[x]) {
          int ly = j.first, ry = j.second;
          if (ly <= y && ry > y) {
            cur[ry].emplace_back(x);
            add(x, 1);
          }
        }
      }
      for (int yy = y + 1; yy <= N; yy++) {
        for (auto j : row[yy]) {
          int res = query(j.second) - query(j.first - 1);
          ans += (res * (res - 1)) >> 1;
        }
        for (auto j : cur[yy]) add(j, -1);
      }
    }
  }
  return !printf("%I64d", ans);
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
inline int read() {
  char ch = getchar();
  int f = 1, x = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f * x;
}
const int N = 10001;
int n, t[N + 1];
vector<int> cur[N + 1];
vector<pair<int, int> > row[N + 1], col[N + 1];
inline void init() {
  memset(t, 0, sizeof(t));
  for (int i = 1; i <= N; i++) cur[i].clear();
}
inline void add(int x, int y) {
  for (; x <= N; x += x & -x) t[x] += y;
}
inline int query(int x) {
  int ans = 0;
  for (; x; x -= x & -x) ans += t[x];
  return ans;
}
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int x1, y1, x2, y2;
    x1 = read() + 5001;
    y1 = read() + 5001;
    x2 = read() + 5001;
    y2 = read() + 5001;
    if (x1 > x2) x1 ^= x2 ^= x1 ^= x2;
    if (y1 > y2) y1 ^= y2 ^= y1 ^= y2;
    if (y1 == y2)
      row[y1].emplace_back(make_pair(x1, x2));
    else
      col[x1].emplace_back(make_pair(y1, y2));
  }
  long long ans = 0;
  for (int y = 1; y <= N; y++) {
    for (auto i : row[y]) {
      init();
      int lx = i.first, rx = i.second;
      for (int x = lx; x <= rx; x++) {
        for (auto j : col[x]) {
          int ly = j.first, ry = j.second;
          if (ly <= y && ry > y) {
            cur[ry].emplace_back(x);
            add(x, 1);
          }
        }
      }
      for (int yy = y + 1; yy <= N; yy++) {
        for (auto j : row[yy]) {
          int res = query(j.second) - query(j.first - 1);
          ans += (res * (res - 1)) >> 1;
        }
        for (auto j : cur[yy]) add(j, -1);
      }
    }
  }
  return !printf("%I64d", ans);
}

```