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
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-12;
const int inf = 2000000000;
const long long int infLL = (long long int)1e18;
long long int MOD = 1000000007;
int MOD1 = 1000000007;
int MOD2 = 1000000009;
inline bool checkBit(long long int n, long long int i) {
  return n & (1LL << i);
}
inline long long int setBit(long long int n, long long int i) {
  return n | (1LL << i);
  ;
}
inline long long int resetBit(long long int n, long long int i) {
  return n & (~(1LL << i));
}
int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long int year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline void normal(long long int &a) {
  a %= MOD;
  (a < 0) && (a += MOD);
}
inline long long int modMul(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a * b) % MOD;
}
inline long long int modAdd(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a + b) % MOD;
}
inline long long int modSub(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long int modPow(long long int b, long long int p) {
  long long int r = 1LL;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1LL;
  }
  return r;
}
inline long long int modDiv(long long int a, long long int b) {
  return modMul(a, modPow(b, MOD - 2));
}
bool comp(const pair<long long int, pair<long long int, long long int> > &p1,
          const pair<long long int, pair<long long int, long long int> > &p2) {
  return p1.first > p2.first;
}
bool comp1(const pair<long long int, long long int> &p1,
           const pair<long long int, long long int> &p2) {
  if (p1.first == p2.first) {
    return p1.second > p2.second;
  }
  return p1.first < p2.first;
}
long long int converter(string a) {
  long long int i, mul = 1, r, t, ans = 0LL;
  if (a.length() == 0) return 0;
  for (i = a.length() - 1; i >= 0; i--) {
    t = a[i] - '0';
    r = t % 10;
    ans += (mul * r);
    mul = mul * 10;
  }
  return ans;
}
int msb(unsigned x) {
  union {
    double a;
    int b[2];
  };
  a = x;
  return (b[1] >> 20) - 1023;
}
const int MAX = (int)1e5 + 5;
vector<long long int> x[MAX];
vector<long long int> y[MAX];
pair<long long int, long long int> p[MAX];
unordered_map<long long int, int> mp[MAX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n, i, j, a, b, ans = 0LL;
  cin >> n;
  for (i = 1; i <= n; ++i) {
    cin >> a >> b;
    p[i] = make_pair(a, b);
    mp[a][b]++;
    x[a].push_back(b);
    y[b].push_back(a);
  }
  for (i = 0; i < MAX; ++i) {
    sort((x[i]).begin(), (x[i]).end());
    sort((y[i]).begin(), (y[i]).end());
  }
  for (i = 1; i <= n; ++i) {
    a = p[i].first;
    b = p[i].second;
    if (x[a].size() <= y[b].size()) {
      for (j = 0; j < x[a].size(); ++j) {
        long long int cury = x[a][j];
        if (cury >= b) break;
        long long int dif = b - cury;
        if (a - dif < 0) continue;
        if (mp[a - dif][b] == 1 && mp[a - dif][cury] == 1) ++ans;
      }
    } else {
      for (j = 0; j < y[b].size(); ++j) {
        long long int curx = y[b][j];
        if (curx >= a) break;
        long long int dif = a - curx;
        if (b - dif < 0) continue;
        if (mp[a][b - dif] == 1 && mp[curx][b - dif] == 1) ++ans;
      }
    }
  }
  cout << ans << '\n';
  return 0;
}
