#include <bits/stdc++.h>
using namespace std;
int W, N, M, X1, Y1, X2, Y2, T[7 * 5005], POS, VAL, S, E;
long long Ans;
priority_queue<pair<int, int> > Q;
struct seg {
  int s, e, x;
} v[5005], h[5005];
bool cond(seg a, seg b) { return a.x < b.x; }
bool cond2(seg a, seg b) { return a.s < b.s; }
void Update(int node, int s, int e) {
  if (s == e) {
    T[node] += VAL;
  } else {
    int mid = (s + e) >> 1;
    if (POS <= mid)
      Update(2 * node, s, mid);
    else
      Update(2 * node + 1, mid + 1, e);
    T[node] = T[2 * node] + T[2 * node + 1];
  }
}
void Query(int node, int s, int e) {
  if (S <= s && e <= E) {
    VAL += T[node];
  } else {
    int mid = (s + e) >> 1;
    if (S <= mid) Query(2 * node, s, mid);
    if (E > mid) Query(2 * node + 1, mid + 1, e);
  }
}
int main() {
  cin >> W;
  for (int i = 1; i <= W; i++) {
    cin >> X1 >> Y1 >> X2 >> Y2;
    if (X1 == X2) {
      h[++M].x = X1;
      h[M].s = min(Y1, Y2);
      h[M].e = max(Y1, Y2);
    } else {
      v[++N].x = Y1;
      v[N].s = min(X1, X2);
      v[N].e = max(X1, X2);
    }
  }
  sort(v + 1, v + 1 + N, cond);
  sort(h + 1, h + 1 + M, cond2);
  for (int i = 1; i <= N; i++) {
    memset(T, 0, sizeof T);
    while (!Q.empty()) Q.pop();
    int pos = 1;
    while (pos <= M && h[pos].s <= v[i].x) {
      POS = h[pos].x;
      VAL = 1;
      Update(1, -5000, 5000);
      Q.push({-h[pos].e, h[pos].x});
      pos++;
    }
    for (int j = i + 1; j <= N; j++) {
      while (!Q.empty() && -Q.top().first < v[j].x) {
        VAL = -1;
        POS = Q.top().second;
        Update(1, -5000, 5000);
        Q.pop();
      }
      VAL = 0;
      S = max(v[i].s, v[j].s);
      E = min(v[i].e, v[j].e);
      if (S <= E) Query(1, -5000, 5000);
      Ans += VAL * (VAL - 1) / 2;
    }
  }
  cout << Ans;
  return 0;
}

### 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;
int W, N, M, X1, Y1, X2, Y2, T[7 * 5005], POS, VAL, S, E;
long long Ans;
priority_queue<pair<int, int> > Q;
struct seg {
  int s, e, x;
} v[5005], h[5005];
bool cond(seg a, seg b) { return a.x < b.x; }
bool cond2(seg a, seg b) { return a.s < b.s; }
void Update(int node, int s, int e) {
  if (s == e) {
    T[node] += VAL;
  } else {
    int mid = (s + e) >> 1;
    if (POS <= mid)
      Update(2 * node, s, mid);
    else
      Update(2 * node + 1, mid + 1, e);
    T[node] = T[2 * node] + T[2 * node + 1];
  }
}
void Query(int node, int s, int e) {
  if (S <= s && e <= E) {
    VAL += T[node];
  } else {
    int mid = (s + e) >> 1;
    if (S <= mid) Query(2 * node, s, mid);
    if (E > mid) Query(2 * node + 1, mid + 1, e);
  }
}
int main() {
  cin >> W;
  for (int i = 1; i <= W; i++) {
    cin >> X1 >> Y1 >> X2 >> Y2;
    if (X1 == X2) {
      h[++M].x = X1;
      h[M].s = min(Y1, Y2);
      h[M].e = max(Y1, Y2);
    } else {
      v[++N].x = Y1;
      v[N].s = min(X1, X2);
      v[N].e = max(X1, X2);
    }
  }
  sort(v + 1, v + 1 + N, cond);
  sort(h + 1, h + 1 + M, cond2);
  for (int i = 1; i <= N; i++) {
    memset(T, 0, sizeof T);
    while (!Q.empty()) Q.pop();
    int pos = 1;
    while (pos <= M && h[pos].s <= v[i].x) {
      POS = h[pos].x;
      VAL = 1;
      Update(1, -5000, 5000);
      Q.push({-h[pos].e, h[pos].x});
      pos++;
    }
    for (int j = i + 1; j <= N; j++) {
      while (!Q.empty() && -Q.top().first < v[j].x) {
        VAL = -1;
        POS = Q.top().second;
        Update(1, -5000, 5000);
        Q.pop();
      }
      VAL = 0;
      S = max(v[i].s, v[j].s);
      E = min(v[i].e, v[j].e);
      if (S <= E) Query(1, -5000, 5000);
      Ans += VAL * (VAL - 1) / 2;
    }
  }
  cout << Ans;
  return 0;
}

```