#include <bits/stdc++.h>
using namespace std;
void read(long long &x) { scanf("%lld", &x); }
void read(long long &x, long long &y) { scanf("%lld%lld", &x, &y); }
void read(long long &x, long long &y, long long &z) {
  scanf("%lld%lld%lld", &x, &y, &z);
}
void print(long long x) { printf("%lld ", x); }
void println(long long x) { printf("%lld\n", x); }
const long long N = 300100;
long long n, x1[N], solai[N], x2[N], y2[N], n1, n2, m, ans, tree[N], calc;
struct item {
  long long k, a, b;
} h[N], v[N], li[N];
vector<item> g[N];
bool cmp(item a, item b) { return a.b > b.b; }
void add(long long pos, long long x) {
  for (long long i = pos; i <= 10000; i = (i | (i + 1))) tree[i] += x;
}
long long sum(long long pos) {
  long long ans = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) ans += tree[pos];
  return ans;
}
void init() {
  for (long long i = 0; i <= 10000; i++) tree[i] = 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> x1[i] >> solai[i] >> x2[i] >> y2[i];
    x1[i] += 5000;
    solai[i] += 5000;
    x2[i] += 5000;
    y2[i] += 5000;
    if (x1[i] == x2[i]) {
      n2++;
      v[n2].k = x1[i];
      v[n2].a = min(solai[i], y2[i]);
      v[n2].b = max(solai[i], y2[i]);
    } else {
      n1++;
      h[n1].k = solai[i];
      h[n1].a = min(x1[i], x2[i]);
      h[n1].b = max(x1[i], x2[i]);
      g[solai[i]].push_back(h[n1]);
    }
  }
  sort(v + 1, v + n2 + 1, &cmp);
  for (long long i = 0; i <= 10000; i++)
    if (!g[i].empty()) {
      m = 0;
      init();
      for (long long j = 1; j <= n2; j++)
        if (v[j].a <= i && v[j].b >= i + 1) li[++m] = v[j], add(v[j].k, 1);
      for (long long j = i + 1; j <= 10000; j++)
        if (!g[j].empty()) {
          while (m && li[m].b < j) add(li[m].k, -1), m--;
          for (item l : g[i])
            for (item r : g[j]) {
              if (l.b <= r.a || r.b <= l.a) continue;
              calc = sum(min(l.b, r.b)) - sum(max(l.a, r.a) - 1);
              ans += calc * (calc - 1) / 2;
            }
        }
    }
  cout << ans << "\n";
}

### Prompt

Please formulate a CPP solution 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;
void read(long long &x) { scanf("%lld", &x); }
void read(long long &x, long long &y) { scanf("%lld%lld", &x, &y); }
void read(long long &x, long long &y, long long &z) {
  scanf("%lld%lld%lld", &x, &y, &z);
}
void print(long long x) { printf("%lld ", x); }
void println(long long x) { printf("%lld\n", x); }
const long long N = 300100;
long long n, x1[N], solai[N], x2[N], y2[N], n1, n2, m, ans, tree[N], calc;
struct item {
  long long k, a, b;
} h[N], v[N], li[N];
vector<item> g[N];
bool cmp(item a, item b) { return a.b > b.b; }
void add(long long pos, long long x) {
  for (long long i = pos; i <= 10000; i = (i | (i + 1))) tree[i] += x;
}
long long sum(long long pos) {
  long long ans = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) ans += tree[pos];
  return ans;
}
void init() {
  for (long long i = 0; i <= 10000; i++) tree[i] = 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> x1[i] >> solai[i] >> x2[i] >> y2[i];
    x1[i] += 5000;
    solai[i] += 5000;
    x2[i] += 5000;
    y2[i] += 5000;
    if (x1[i] == x2[i]) {
      n2++;
      v[n2].k = x1[i];
      v[n2].a = min(solai[i], y2[i]);
      v[n2].b = max(solai[i], y2[i]);
    } else {
      n1++;
      h[n1].k = solai[i];
      h[n1].a = min(x1[i], x2[i]);
      h[n1].b = max(x1[i], x2[i]);
      g[solai[i]].push_back(h[n1]);
    }
  }
  sort(v + 1, v + n2 + 1, &cmp);
  for (long long i = 0; i <= 10000; i++)
    if (!g[i].empty()) {
      m = 0;
      init();
      for (long long j = 1; j <= n2; j++)
        if (v[j].a <= i && v[j].b >= i + 1) li[++m] = v[j], add(v[j].k, 1);
      for (long long j = i + 1; j <= 10000; j++)
        if (!g[j].empty()) {
          while (m && li[m].b < j) add(li[m].k, -1), m--;
          for (item l : g[i])
            for (item r : g[j]) {
              if (l.b <= r.a || r.b <= l.a) continue;
              calc = sum(min(l.b, r.b)) - sum(max(l.a, r.a) - 1);
              ans += calc * (calc - 1) / 2;
            }
        }
    }
  cout << ans << "\n";
}

```