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
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF425D extends PrintWriter {
	CF425D() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF425D o = new CF425D(); o.main(); o.flush();
	}

	static final int K = 200, Y = 100000;
	static class P {
		int x, y;
		P(int x, int y) {
			this.x = x; this.y = y;
		}
	}
	static Random rand = new Random();
	static class V {
		V l, r;
		long key;
		int x;
		V(long key) {
			this.key = key;
			x = rand.nextInt();
		}
	}
	V v_, l_, r_;
	void split(V v, long key) {
		if (v == null) {
			v_ = l_ = r_ = null;
		} else if (v.key < key) {
			split(v.r, key);
			v.r = l_; l_ = v;
		} else if (v.key > key) {
			split(v.l, key);
			v.l = r_; r_ = v;
		} else {
			v_ = v; l_ = v.l; r_ = v.r;
			v.l = v.r = null;
		}
	}
	V merge(V u, V v) {
		if (u == null)
			return v;
		if (v == null)
			return u;
		if (u.x < v.x) {
			u.r = merge(u.r, v);
			return u;
		} else {
			v.l = merge(u, v.l);
			return v;
		}
	}
	boolean has(long key) {
		split(v_, key);
		boolean ans = v_ != null;
		v_ = merge(l_, merge(v_, r_));
		return ans;
	}
	void add(long key) {
		split(v_, key);
		if (v_ == null)
			v_ = new V(key);
		v_ = merge(l_, merge(v_, r_));
	}
	long compose(int x, int y) {
		return (long) x * (Y + 1) + y;
	}
	void main() {
		int n = sc.nextInt();
		P[] pp = new P[n];
		for (int i = 0; i < n; i++) {
			int x = sc.nextInt();
			int y = sc.nextInt();
			pp[i] = new P(x, y);
			add(compose(x, y));
		}
		int[] ll = new int[Y + 1];
		int[] rr = new int[Y + 1]; Arrays.fill(rr, -1);
		Arrays.sort(pp, (p, q) -> p.y != q.y ? p.y - q.y : p.x - q.x);
		int[] yy = new int[n]; int cnt = 0;
		for (int i = 0, j; i < n; i = j) {
			int y = pp[i].y;
			j = i + 1;
			while (j < n && pp[j].y == y)
				j++;
			ll[y] = i; rr[y] = j;
			if (j - i > K)
				yy[cnt++] = y;
		}
		int ans = 0;
		for (int h = 0; h < cnt; h++) {
			int y = yy[h], l = ll[y], r = rr[y];
			for (int h_ = h + 1; h_ < cnt; h_++) {
				int y_ = yy[h_], l_ = ll[y_], r_ = rr[y_];
				for (int i = l, j = l, i_ = l_, j_ = l_; i < r && j < r && i_ < r_ && j_ < r_; i++) {
					int x = pp[i].x, x_ = x + y_ - y;
					while (j < r && pp[j].x < x_)
						j++;
					if (j < r && pp[j].x == x_) {
						while (i_ < r_ && pp[i_].x < x)
							i_++;
						if (i_ < r_ && pp[i_].x == x) {
							while (j_ < r_ && pp[j_].x < x_)
								j_++;
							if (j_ < r_ && pp[j_].x == x_)
								ans++;
						}
					}
				}
			}
		}
		for (int y = 0; y <= Y; y++) {
			int l = ll[y], r = rr[y];
			if (r - l > K)
				continue;
			for (int i = l; i < r; i++)
				for (int j = i + 1; j < r; j++) {
					int x = pp[i].x, x_ = pp[j].x, y_, l_, r_;
					y_ = y - (x_ - x);
					if (y_ >= 0 && rr[y_] - ll[y_] > K && has(compose(x, y_)) && has(compose(x_, y_)))
						ans++;
					y_ = y + (x_ - x);
					if (y_ <= Y && rr[y_] - ll[y_] > 0 && has(compose(x, y_)) && has(compose(x_, y_)))
						ans++;
				}
		}
		println(ans);
	}
}
