#include <bits/stdc++.h>
using namespace std;
const long long INF = 0x3f3f3f3f;
const int N = 1e4 + 10;
const long long mod = 1e9 + 7;
const bool debug = false;
int mid = 5005;
pair<pair<int, int>, int> ho[N], ve[N];
struct BITstree {
  int node[N];
  void init() { memset(node, 0, sizeof node); }
  int sum(int x) {
    int ret = 0;
    for (; x; x -= (x & (-x))) ret += node[x];
    return ret;
  }
  void up(int x, int n, int d) {
    for (; x <= n; x += (x & (-x))) node[x] += d;
  }
} bit;
bool cmp(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  return x.second < y.second;
}
bool cmp2(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  return x.first.first < y.first.first;
}
int main() {
  int n, tot1 = 0, tot2 = 0;
  scanf("%d", &n);
  for (int i = 1, x1, x2, y1, y2; i <= n; ++i) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += mid, x2 += mid, y1 += mid, y2 += mid;
    if (x1 == x2) {
      if (y1 > y2) swap(y1, y2);
      ho[++tot1] = {{y1, y2}, x1};
    } else {
      if (x1 > x2) swap(x1, x2);
      ve[++tot2] = {{x1, x2}, y1};
    }
  }
  long long ans = 0;
  sort(ho + 1, ho + tot1 + 1, cmp);
  sort(ve + 1, ve + tot2 + 1, cmp2);
  int mx = N - 1;
  for (int i = 2; i <= tot1; ++i) {
    bit.init();
    for (int j = 1, k = 1, si; j < i; ++j) {
      while (k <= tot2 && ve[k].first.first <= ho[j].second) {
        if (ve[k].first.second >= ho[i].second &&
            ve[k].second >= ho[i].first.first &&
            ve[k].second <= ho[i].first.second)
          bit.up(ve[k].second, mx, 1);
        k++;
      }
      si = (bit.sum(ho[j].first.second) - bit.sum(ho[j].first.first - 1));
      ans += si * 1ll * (si - 1) / 2;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 long long INF = 0x3f3f3f3f;
const int N = 1e4 + 10;
const long long mod = 1e9 + 7;
const bool debug = false;
int mid = 5005;
pair<pair<int, int>, int> ho[N], ve[N];
struct BITstree {
  int node[N];
  void init() { memset(node, 0, sizeof node); }
  int sum(int x) {
    int ret = 0;
    for (; x; x -= (x & (-x))) ret += node[x];
    return ret;
  }
  void up(int x, int n, int d) {
    for (; x <= n; x += (x & (-x))) node[x] += d;
  }
} bit;
bool cmp(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  return x.second < y.second;
}
bool cmp2(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  return x.first.first < y.first.first;
}
int main() {
  int n, tot1 = 0, tot2 = 0;
  scanf("%d", &n);
  for (int i = 1, x1, x2, y1, y2; i <= n; ++i) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += mid, x2 += mid, y1 += mid, y2 += mid;
    if (x1 == x2) {
      if (y1 > y2) swap(y1, y2);
      ho[++tot1] = {{y1, y2}, x1};
    } else {
      if (x1 > x2) swap(x1, x2);
      ve[++tot2] = {{x1, x2}, y1};
    }
  }
  long long ans = 0;
  sort(ho + 1, ho + tot1 + 1, cmp);
  sort(ve + 1, ve + tot2 + 1, cmp2);
  int mx = N - 1;
  for (int i = 2; i <= tot1; ++i) {
    bit.init();
    for (int j = 1, k = 1, si; j < i; ++j) {
      while (k <= tot2 && ve[k].first.first <= ho[j].second) {
        if (ve[k].first.second >= ho[i].second &&
            ve[k].second >= ho[i].first.first &&
            ve[k].second <= ho[i].first.second)
          bit.up(ve[k].second, mx, 1);
        k++;
      }
      si = (bit.sum(ho[j].first.second) - bit.sum(ho[j].first.first - 1));
      ans += si * 1ll * (si - 1) / 2;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```