#include <bits/stdc++.h>
const double Pi = 3.14159265;
const int N = 1e4 + 20;
const unsigned long long base = 163;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-10;
const double pi = acos(-1);
using namespace std;
int n, m;
int val[N];
void add(int x, int w) {
  for (; x < N; x += x & -x) {
    val[x] += w;
  }
}
int sum(int x) {
  int ans = 0;
  for (; x; x -= x & -x) {
    ans += val[x];
  }
  return ans;
}
pair<int, pair<int, int> > row[N], col[N];
int st[N], sz = 0;
int main() {
  scanf("%d", &n);
  int r = 0, c = 0;
  for (int i = 1; i <= n; ++i) {
    int a1, b1, a2, b2;
    scanf("%d%d%d%d", &a1, &b1, &a2, &b2);
    a1 += 5005;
    b1 += 5005;
    a2 += 5005;
    b2 += 5005;
    if (a1 == a2) {
      if (b1 > b2) swap(b1, b2);
      col[++c] = {a1, {b1, b2}};
    } else {
      if (a1 > a2) swap(a1, a2);
      row[++r] = {a2, {a1, b1}};
    }
  }
  sort(row + 1, row + 1 + r);
  sort(col + 1, col + 1 + c);
  long long ans = 0;
  for (int i = 1; i <= c; ++i) {
    int rr = r + 1;
    memset(val, 0, sizeof(val));
    sz = 0;
    for (int j = 1; j <= r; ++j) {
      if (row[j].second.first <= col[i].first)
        add(row[j].second.second, 1), st[++sz] = j;
    }
    int pos = 1;
    for (int j = i + 1; j <= c; ++j) {
      while (row[st[pos]].first < col[j].first && pos <= sz)
        add(row[st[pos]].second.second, -1), pos++;
      int MIN = min(col[i].second.second, col[j].second.second);
      int MAX = max(col[i].second.first, col[j].second.first);
      if (MAX > MIN) continue;
      int now = sum(MIN) - sum(MAX - 1);
      ans += 1LL * now * (now - 1) / 2;
    }
  }
  cout << ans << endl;
  return 0;
}

### 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>
const double Pi = 3.14159265;
const int N = 1e4 + 20;
const unsigned long long base = 163;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-10;
const double pi = acos(-1);
using namespace std;
int n, m;
int val[N];
void add(int x, int w) {
  for (; x < N; x += x & -x) {
    val[x] += w;
  }
}
int sum(int x) {
  int ans = 0;
  for (; x; x -= x & -x) {
    ans += val[x];
  }
  return ans;
}
pair<int, pair<int, int> > row[N], col[N];
int st[N], sz = 0;
int main() {
  scanf("%d", &n);
  int r = 0, c = 0;
  for (int i = 1; i <= n; ++i) {
    int a1, b1, a2, b2;
    scanf("%d%d%d%d", &a1, &b1, &a2, &b2);
    a1 += 5005;
    b1 += 5005;
    a2 += 5005;
    b2 += 5005;
    if (a1 == a2) {
      if (b1 > b2) swap(b1, b2);
      col[++c] = {a1, {b1, b2}};
    } else {
      if (a1 > a2) swap(a1, a2);
      row[++r] = {a2, {a1, b1}};
    }
  }
  sort(row + 1, row + 1 + r);
  sort(col + 1, col + 1 + c);
  long long ans = 0;
  for (int i = 1; i <= c; ++i) {
    int rr = r + 1;
    memset(val, 0, sizeof(val));
    sz = 0;
    for (int j = 1; j <= r; ++j) {
      if (row[j].second.first <= col[i].first)
        add(row[j].second.second, 1), st[++sz] = j;
    }
    int pos = 1;
    for (int j = i + 1; j <= c; ++j) {
      while (row[st[pos]].first < col[j].first && pos <= sz)
        add(row[st[pos]].second.second, -1), pos++;
      int MIN = min(col[i].second.second, col[j].second.second);
      int MAX = max(col[i].second.first, col[j].second.first);
      if (MAX > MIN) continue;
      int now = sum(MIN) - sum(MAX - 1);
      ans += 1LL * now * (now - 1) / 2;
    }
  }
  cout << ans << endl;
  return 0;
}

```