#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
const int INF = (1 << 29);
const int mod = 1000007;
int a, b, c, d, e, f, g, h, i, j, k;
pair<int, int> G[N];
vector<long long> hav[2][1000025];
vector<int> GG[N], Big;
int tot, type[N];
int max_x = -INF, max_y = -INF, min_x = INF, min_y = INF;
int an;
inline long long Num(int x, int y) {
  return 2000000LL * ((long long)x + 2000000 - 1) + ((long long)y + 2000000);
}
inline bool Hash(vector<long long> uu[], int x, int y) {
  long long tt = Num(x, y);
  for (int ii = 0; ii < uu[tt % mod].size(); ii++)
    if (uu[tt % mod][ii] == tt) return true;
  return false;
}
inline void Check1(int x, int y) {
  int delta = G[y].second - G[x].second;
  if (G[x].first - delta >= min_x &&
      Hash(hav[0], G[x].first - delta, G[x].second) &&
      Hash(hav[0], G[x].first - delta, G[y].second))
    an++;
  if (G[x].first - delta >= min_x &&
      Hash(hav[1], G[x].first - delta, G[x].second) &&
      Hash(hav[1], G[x].first - delta, G[y].second))
    an++;
  if (G[x].first + delta <= max_x &&
      Hash(hav[1], G[x].first + delta, G[x].second) &&
      Hash(hav[1], G[x].first + delta, G[y].second))
    an++;
}
inline void Check2(int x, int y) {
  int delta = G[y].first - G[x].first;
  if (G[x].second - delta >= min_y &&
      Hash(hav[1], G[x].first, G[x].second - delta) &&
      Hash(hav[1], G[y].first, G[y].second - delta))
    an++;
}
int main() {
  cin >> a;
  for (i = 1; i <= a; i++) {
    scanf("%d%d", &G[i].first, &G[i].second);
    max_x = max(max_x, G[i].first);
    max_y = max(max_y, G[i].second);
    min_x = min(min_x, G[i].first);
    min_y = min(min_y, G[i].second);
  }
  sort(G + 1, G + a + 1);
  for (i = 1; i <= a; i++) {
    if (i == 1 || G[i].first != G[i - 1].first) tot++;
    GG[tot].push_back(i);
  }
  int siz = (int)sqrt((double)a);
  for (i = 1; i <= tot; i++) {
    int vv = 1;
    if (GG[i].size() <= siz)
      vv = 0;
    else
      Big.push_back(i);
    for (j = 0; j < GG[i].size(); j++) {
      long long tt = Num(G[GG[i][j]].first, G[GG[i][j]].second);
      hav[vv][tt % mod].push_back(tt);
    }
  }
  int Cnt = 0;
  for (i = 1; i <= tot; i++)
    if (GG[i].size() <= siz) {
      Cnt += GG[i].size() * GG[i].size();
      for (j = 0; j < GG[i].size(); j++)
        for (k = j + 1; k < GG[i].size(); k++) Check1(GG[i][j], GG[i][k]);
    }
  for (i = 0; i < Big.size(); i++)
    for (j = i + 1; j < Big.size(); j++) {
      c = Big[i], d = Big[j];
      vector<pair<int, int> > u;
      for (k = 0; k < GG[d].size(); k++)
        u.push_back(make_pair(G[GG[d][k]].second, GG[d][k]));
      int pos = 0;
      for (k = 0; k < GG[c].size(); k++) {
        while (pos < u.size() && G[GG[c][k]].second > u[pos].first) pos++;
        if (pos == u.size()) break;
        if (G[GG[c][k]].second < u[pos].first) continue;
        Check2(GG[c][k], u[pos].second);
      }
    }
  cout << an << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

Sereja has painted n distinct points on the plane. The coordinates of each point are integers. Now he is wondering: how many squares are there with sides parallel to the coordinate axes and with points painted in all its four vertexes? Help him, calculate this number.

Input

The first line contains integer n (1 ≤ n ≤ 105). Each of the next n lines contains two integers xi, yi (0 ≤ xi, yi ≤ 105), the integers represent the coordinates of the i-th point. It is guaranteed that all the given points are distinct.

Output

In a single line print the required number of squares.

Examples

Input

5
0 0
0 2
2 0
2 2
1 1


Output

1


Input

9
0 0
1 1
2 2
0 1
1 0
0 2
2 0
1 2
2 1


Output

5

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
const int INF = (1 << 29);
const int mod = 1000007;
int a, b, c, d, e, f, g, h, i, j, k;
pair<int, int> G[N];
vector<long long> hav[2][1000025];
vector<int> GG[N], Big;
int tot, type[N];
int max_x = -INF, max_y = -INF, min_x = INF, min_y = INF;
int an;
inline long long Num(int x, int y) {
  return 2000000LL * ((long long)x + 2000000 - 1) + ((long long)y + 2000000);
}
inline bool Hash(vector<long long> uu[], int x, int y) {
  long long tt = Num(x, y);
  for (int ii = 0; ii < uu[tt % mod].size(); ii++)
    if (uu[tt % mod][ii] == tt) return true;
  return false;
}
inline void Check1(int x, int y) {
  int delta = G[y].second - G[x].second;
  if (G[x].first - delta >= min_x &&
      Hash(hav[0], G[x].first - delta, G[x].second) &&
      Hash(hav[0], G[x].first - delta, G[y].second))
    an++;
  if (G[x].first - delta >= min_x &&
      Hash(hav[1], G[x].first - delta, G[x].second) &&
      Hash(hav[1], G[x].first - delta, G[y].second))
    an++;
  if (G[x].first + delta <= max_x &&
      Hash(hav[1], G[x].first + delta, G[x].second) &&
      Hash(hav[1], G[x].first + delta, G[y].second))
    an++;
}
inline void Check2(int x, int y) {
  int delta = G[y].first - G[x].first;
  if (G[x].second - delta >= min_y &&
      Hash(hav[1], G[x].first, G[x].second - delta) &&
      Hash(hav[1], G[y].first, G[y].second - delta))
    an++;
}
int main() {
  cin >> a;
  for (i = 1; i <= a; i++) {
    scanf("%d%d", &G[i].first, &G[i].second);
    max_x = max(max_x, G[i].first);
    max_y = max(max_y, G[i].second);
    min_x = min(min_x, G[i].first);
    min_y = min(min_y, G[i].second);
  }
  sort(G + 1, G + a + 1);
  for (i = 1; i <= a; i++) {
    if (i == 1 || G[i].first != G[i - 1].first) tot++;
    GG[tot].push_back(i);
  }
  int siz = (int)sqrt((double)a);
  for (i = 1; i <= tot; i++) {
    int vv = 1;
    if (GG[i].size() <= siz)
      vv = 0;
    else
      Big.push_back(i);
    for (j = 0; j < GG[i].size(); j++) {
      long long tt = Num(G[GG[i][j]].first, G[GG[i][j]].second);
      hav[vv][tt % mod].push_back(tt);
    }
  }
  int Cnt = 0;
  for (i = 1; i <= tot; i++)
    if (GG[i].size() <= siz) {
      Cnt += GG[i].size() * GG[i].size();
      for (j = 0; j < GG[i].size(); j++)
        for (k = j + 1; k < GG[i].size(); k++) Check1(GG[i][j], GG[i][k]);
    }
  for (i = 0; i < Big.size(); i++)
    for (j = i + 1; j < Big.size(); j++) {
      c = Big[i], d = Big[j];
      vector<pair<int, int> > u;
      for (k = 0; k < GG[d].size(); k++)
        u.push_back(make_pair(G[GG[d][k]].second, GG[d][k]));
      int pos = 0;
      for (k = 0; k < GG[c].size(); k++) {
        while (pos < u.size() && G[GG[c][k]].second > u[pos].first) pos++;
        if (pos == u.size()) break;
        if (G[GG[c][k]].second < u[pos].first) continue;
        Check2(GG[c][k], u[pos].second);
      }
    }
  cout << an << endl;
  return 0;
}

```