#include <bits/stdc++.h>
using namespace std;
void B();
int main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  B();
}
class segmen {
 public:
  long long x1, y1, x2, y2;
  segmen(long long xx1, long long yy1, long long xx2, long long yy2)
      : x1{xx1}, x2{xx2}, y1{yy1}, y2{yy2} {}
  segmen() {}
};
vector<segmen> hor, ver;
const long long N = 10010;
const long long shift = 5001;
bool compar(segmen A, segmen B) { return A.y1 <= B.y1; }
long long T[N];
void upd(long long x, long long val) {
  for (; x < N; x += (x & (-x))) T[x] += val;
}
long long F(long long x) {
  long long ret = 0;
  for (; x; x -= (x & (-x))) ret += T[x];
  return ret;
}
long long gett(long long l, long long r) { return F(r) - F(l - 1); }
vector<pair<long long, long long> > H[N];
vector<pair<long long, long long> > V[N];
void B() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long x1, x2, y1, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += shift;
    x2 += shift;
    y1 += shift;
    y2 += shift;
    if (y1 == y2) {
      H[y1].push_back({min(x1, x2), max(x1, x2)});
    }
    if (x1 == x2) {
      V[x1].push_back({min(y1, y2), max(y1, y2)});
    }
  }
  long long ans = 0;
  vector<long long> endy[N];
  for (long long i = 1; i < N; i++) {
    for (auto A : H[i]) {
      long long xl = A.first;
      long long xr = A.second;
      long long maxn = 0;
      for (long long j = xl; j <= xr; j++) {
        for (auto B : V[j]) {
          if (B.first <= i && B.second > i) {
            upd(j, 1);
            endy[B.second].push_back(j);
            maxn = max(maxn, B.second);
          }
        }
      }
      for (long long j = i + 1; j <= maxn; j++) {
        for (auto B : H[j]) {
          long long ret = gett(B.first, B.second);
          ans += (ret * (ret - 1)) >> 1;
        }
        for (auto B : endy[j]) {
          upd(B, -1);
        }
        endy[j].clear();
      }
    }
  }
  cout << ans << '\n';
}

### Prompt

Please provide a Cpp coded solution 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;
void B();
int main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  B();
}
class segmen {
 public:
  long long x1, y1, x2, y2;
  segmen(long long xx1, long long yy1, long long xx2, long long yy2)
      : x1{xx1}, x2{xx2}, y1{yy1}, y2{yy2} {}
  segmen() {}
};
vector<segmen> hor, ver;
const long long N = 10010;
const long long shift = 5001;
bool compar(segmen A, segmen B) { return A.y1 <= B.y1; }
long long T[N];
void upd(long long x, long long val) {
  for (; x < N; x += (x & (-x))) T[x] += val;
}
long long F(long long x) {
  long long ret = 0;
  for (; x; x -= (x & (-x))) ret += T[x];
  return ret;
}
long long gett(long long l, long long r) { return F(r) - F(l - 1); }
vector<pair<long long, long long> > H[N];
vector<pair<long long, long long> > V[N];
void B() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long x1, x2, y1, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += shift;
    x2 += shift;
    y1 += shift;
    y2 += shift;
    if (y1 == y2) {
      H[y1].push_back({min(x1, x2), max(x1, x2)});
    }
    if (x1 == x2) {
      V[x1].push_back({min(y1, y2), max(y1, y2)});
    }
  }
  long long ans = 0;
  vector<long long> endy[N];
  for (long long i = 1; i < N; i++) {
    for (auto A : H[i]) {
      long long xl = A.first;
      long long xr = A.second;
      long long maxn = 0;
      for (long long j = xl; j <= xr; j++) {
        for (auto B : V[j]) {
          if (B.first <= i && B.second > i) {
            upd(j, 1);
            endy[B.second].push_back(j);
            maxn = max(maxn, B.second);
          }
        }
      }
      for (long long j = i + 1; j <= maxn; j++) {
        for (auto B : H[j]) {
          long long ret = gett(B.first, B.second);
          ans += (ret * (ret - 1)) >> 1;
        }
        for (auto B : endy[j]) {
          upd(B, -1);
        }
        endy[j].clear();
      }
    }
  }
  cout << ans << '\n';
}

```