#include <bits/stdc++.h>
using namespace std;
const int maxn = 5000 + 10;
long long Ans;
int tot, n, m, val[maxn << 2], cnt;
struct data {
  int x, l, r;
  data() {}
  data(int _x, int _l, int _r) : x(_x), l(_l), r(_r) {}
} a[maxn], b[maxn];
inline bool cmp(data a, data b) {
  if (a.x != b.x) return a.x < b.x;
  if (a.l != b.l) return a.l < b.l;
  return a.r < b.r;
}
struct BIT {
  int c[maxn << 3];
  inline int lowbit(int k) { return k & -k; }
  inline void init() { memset(c, 0, sizeof(c)); }
  inline void add(int x, int k) {
    for (int i = x; i <= cnt; i += lowbit(i)) c[i] += k;
  }
  inline int query(int x) {
    if (x <= 0) return 0;
    int tmp = 0;
    for (int i = x; i; i -= lowbit(i)) tmp += c[i];
    return tmp;
  }
  inline int query(int l, int r) {
    if (l > r) return 0;
    return query(r) - query(l - 1);
  }
} B;
vector<int> g[maxn << 1], q[maxn << 1];
inline long long C2(long long x) { return (long long)x * (x - 1) / 2; }
inline void calc(int x, int y) {
  int Sz1 = g[x].size(), Sz2 = g[y].size();
  int Sz = q[x].size();
  for (int i = 0; i < Sz; i++) {
    B.add(q[x][i], 1);
  }
  long long now = 0;
  for (int i = 0; i < Sz1; i++) {
    for (int j = 0; j < Sz2; j++) {
      int id1 = g[x][i], id2 = g[y][j];
      int l = max(a[id1].l, a[id2].l);
      int r = min(a[id1].r, a[id2].r);
      if (l > r) continue;
      long long tmp = B.query(l, r);
      now += C2(tmp);
    }
  }
  Ans += now;
}
inline void work() {
  for (int i = 1; i <= n; i++) g[a[i].x].push_back(i);
  for (int i = 2; i <= cnt; i++) {
    int Sz = g[i].size();
    if (!Sz) continue;
    B.init();
    for (int j = 1; j <= m; j++) {
      if (b[j].r < i) continue;
      q[b[j].l].push_back(b[j].x);
    }
    for (int j = 1; j < i; j++) {
      calc(j, i);
    }
    for (int j = 1; j <= m; j++) {
      if (b[j].r < i) continue;
      B.add(b[j].x, -1);
      vector<int>().swap(q[b[j].l]);
    }
  }
}
int main() {
  scanf("%d", &tot);
  n = m = cnt = 0;
  for (int i = 1; i <= tot; i++) {
    int x1, y1, x2, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2 && y1 != y2)
      a[++n] = data(x1, y1, y2);
    else
      b[++m] = data(y1, x1, x2);
    val[++cnt] = x1;
    val[++cnt] = x2;
    val[++cnt] = y1;
    val[++cnt] = y2;
  }
  sort(val + 1, val + cnt + 1);
  cnt = unique(val + 1, val + cnt + 1) - val - 1;
  for (int i = 1; i <= n; i++) {
    a[i].x = lower_bound(val + 1, val + cnt + 1, a[i].x) - val;
    a[i].l = lower_bound(val + 1, val + cnt + 1, a[i].l) - val;
    a[i].r = lower_bound(val + 1, val + cnt + 1, a[i].r) - val;
  }
  for (int i = 1; i <= m; i++) {
    b[i].x = lower_bound(val + 1, val + cnt + 1, b[i].x) - val;
    b[i].l = lower_bound(val + 1, val + cnt + 1, b[i].l) - val;
    b[i].r = lower_bound(val + 1, val + cnt + 1, b[i].r) - val;
  }
  Ans = 0;
  work();
  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 int maxn = 5000 + 10;
long long Ans;
int tot, n, m, val[maxn << 2], cnt;
struct data {
  int x, l, r;
  data() {}
  data(int _x, int _l, int _r) : x(_x), l(_l), r(_r) {}
} a[maxn], b[maxn];
inline bool cmp(data a, data b) {
  if (a.x != b.x) return a.x < b.x;
  if (a.l != b.l) return a.l < b.l;
  return a.r < b.r;
}
struct BIT {
  int c[maxn << 3];
  inline int lowbit(int k) { return k & -k; }
  inline void init() { memset(c, 0, sizeof(c)); }
  inline void add(int x, int k) {
    for (int i = x; i <= cnt; i += lowbit(i)) c[i] += k;
  }
  inline int query(int x) {
    if (x <= 0) return 0;
    int tmp = 0;
    for (int i = x; i; i -= lowbit(i)) tmp += c[i];
    return tmp;
  }
  inline int query(int l, int r) {
    if (l > r) return 0;
    return query(r) - query(l - 1);
  }
} B;
vector<int> g[maxn << 1], q[maxn << 1];
inline long long C2(long long x) { return (long long)x * (x - 1) / 2; }
inline void calc(int x, int y) {
  int Sz1 = g[x].size(), Sz2 = g[y].size();
  int Sz = q[x].size();
  for (int i = 0; i < Sz; i++) {
    B.add(q[x][i], 1);
  }
  long long now = 0;
  for (int i = 0; i < Sz1; i++) {
    for (int j = 0; j < Sz2; j++) {
      int id1 = g[x][i], id2 = g[y][j];
      int l = max(a[id1].l, a[id2].l);
      int r = min(a[id1].r, a[id2].r);
      if (l > r) continue;
      long long tmp = B.query(l, r);
      now += C2(tmp);
    }
  }
  Ans += now;
}
inline void work() {
  for (int i = 1; i <= n; i++) g[a[i].x].push_back(i);
  for (int i = 2; i <= cnt; i++) {
    int Sz = g[i].size();
    if (!Sz) continue;
    B.init();
    for (int j = 1; j <= m; j++) {
      if (b[j].r < i) continue;
      q[b[j].l].push_back(b[j].x);
    }
    for (int j = 1; j < i; j++) {
      calc(j, i);
    }
    for (int j = 1; j <= m; j++) {
      if (b[j].r < i) continue;
      B.add(b[j].x, -1);
      vector<int>().swap(q[b[j].l]);
    }
  }
}
int main() {
  scanf("%d", &tot);
  n = m = cnt = 0;
  for (int i = 1; i <= tot; i++) {
    int x1, y1, x2, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2 && y1 != y2)
      a[++n] = data(x1, y1, y2);
    else
      b[++m] = data(y1, x1, x2);
    val[++cnt] = x1;
    val[++cnt] = x2;
    val[++cnt] = y1;
    val[++cnt] = y2;
  }
  sort(val + 1, val + cnt + 1);
  cnt = unique(val + 1, val + cnt + 1) - val - 1;
  for (int i = 1; i <= n; i++) {
    a[i].x = lower_bound(val + 1, val + cnt + 1, a[i].x) - val;
    a[i].l = lower_bound(val + 1, val + cnt + 1, a[i].l) - val;
    a[i].r = lower_bound(val + 1, val + cnt + 1, a[i].r) - val;
  }
  for (int i = 1; i <= m; i++) {
    b[i].x = lower_bound(val + 1, val + cnt + 1, b[i].x) - val;
    b[i].l = lower_bound(val + 1, val + cnt + 1, b[i].l) - val;
    b[i].r = lower_bound(val + 1, val + cnt + 1, b[i].r) - val;
  }
  Ans = 0;
  work();
  printf("%lld\n", Ans);
  return 0;
}

```