Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen on a plane, the i-th watchman is located at point (xi, yi).

They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen i and j to be |xi - xj| + |yi - yj|. Daniel, as an ordinary person, calculates the distance using the formula <image>.

The success of the operation relies on the number of pairs (i, j) (1 ≤ i < j ≤ n), such that the distance between watchman i and watchmen j calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.

Input

The first line of the input contains the single integer n (1 ≤ n ≤ 200 000) — the number of watchmen.

Each of the following n lines contains two integers xi and yi (|xi|, |yi| ≤ 109).

Some positions may coincide.

Output

Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.

Examples

Input

3
1 1
7 5
1 5


Output

2


Input

6
0 0
0 1
0 2
-1 1
0 1
1 1


Output

11

Note

In the first sample, the distance between watchman 1 and watchman 2 is equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and <image> for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances.
    import java.util.*;
    import java.lang.Math; 
    import java.util.Arrays;
    import java.util.Scanner;
    import java.util.stream.IntStream;
    import java.io.OutputStream;
    import java.io.IOException;
    import java.io.InputStream;
    import java.io.PrintWriter;
    import java.util.Arrays;
     
    public class Main {
         static int oo = (int)1e9;
       	public static void main(String[] args) {
    	   InputStream inputStream = System.in;
            OutputStream outputStream = System.out;
            Scanner in = new Scanner(inputStream);
            PrintWriter out = new PrintWriter(outputStream);
            TaskA solver = new TaskA();
            solver.solve(1, in, out);
            out.close();
    	}
    	
    	static class TaskA {
      public void solve(int testNumber, Scanner sc, PrintWriter out) {
            int n = sc.nextInt();
		HashMap<Integer,Integer> x = new HashMap<Integer,Integer>();
		HashMap<Integer,Integer> y = new HashMap<Integer,Integer>();
		HashMap<String,Integer> xy = new HashMap<String,Integer>();
		
		long ans =0;
		while(n-->0)
		{
			int a = sc.nextInt();
			int b = sc.nextInt();
			if (x.containsKey(a))
			{
				ans+=x.get(a);
				x.put(a, 1+x.get(a));
			}
			else
				x.put(a,1);
			
			if (y.containsKey(b))
			{
				ans+=y.get(b);
				y.put(b, 1+y.get(b));
			}
			else
				y.put(b,1);
			
			if (xy.containsKey(a+" "+b))
			{
				ans-=xy.get(a+" "+b);
				xy.put(a+" "+b,1+ xy.get(a+" "+b));
			}
			else
				xy.put(a+" "+b, 1);
			
			
		}
	          out.println(ans);
	
       }
    	}     
    	  	
    public int factorial(int n) {
            int fact = 1;
            int i = 1;
           while(i <= n) {
             fact *= i;
             i++;
           }
            return fact;
    }
     
  
    public static long gcd(long x,long y)
    {
    	if(x%y==0)
    		return y;
    	else
    		return gcd(y,x%y);
    }
    public static int gcd(int x,int y)
    {
    	if(x%y==0)
    		return y;
    	else 
    		return gcd(y,x%y);
    }
    public static int abs(int a,int b)
    {
    	return (int)Math.abs(a-b);
    }
    public static long abs(long a,long b)
    {
    	return (long)Math.abs(a-b);
    }
    public static int max(int a,int b)
    {
    	if(a>b)
    		return a;
    	else
    		return b;
    }
    public static int min(int a,int b)
    {
    	if(a>b)
    		return b;
    	else 
    		return a;
    }
    public static long max(long a,long b)
    {
    	if(a>b)
    		return a;
    	else
    		return b;
    }
    public static long min(long a,long b)
    {
    	if(a>b)
    		return b;
    	else 
    		return a;
    }
     
     
    public static long pow(long n,long p,long m)
    {
    	 long  result = 1;
    	  if(p==0)
    	    return 1;
    	if (p==1)
    	    return n;
    	while(p!=0)
    	{
    	    if(p%2==1)
    	        result *= n;
    	    if(result>=m)
    	    result%=m;
    	    p >>=1;
    	    n*=n;
    	    if(n>=m)
    	    n%=m;
    	}
    	return result;
    }
    public static long pow(long n,long p)
    {
    	long  result = 1;
    	  if(p==0)
    	    return 1;
    	if (p==1)
    	    return n;
    	while(p!=0)
    	{
    	    if(p%2==1)
    	        result *= n;	    
    	    p >>=1;
    	    n*=n;	    
    	}
    	return result;
     
    }
     
    static long sort(int a[]){  
         int n=a.length;
    	int b[]=new int[n];	
    	return mergeSort(a,b,0,n-1);
         
    }
     
    static long mergeSort(int a[],int b[],long left,long right){
         long c=0;
         if(left<right){   
          long mid=left+(right-left)/2;
    	 c= mergeSort(a,b,left,mid);
    	 c+=mergeSort(a,b,mid+1,right);
    	 c+=merge(a,b,left,mid+1,right); 
         }	
    	return c;	 
    }
    static long merge(int a[],int  b[],long left,long mid,long right){
         long c=0;int i=(int)left;int j=(int)mid; int k=(int)left;
         while(i<=(int)mid-1&&j<=(int)right){ 
              if(a[i]<=a[j]){
                   b[k++]=a[i++]; 
              }
              else{ 
                   b[k++]=a[j++];c+=mid-i;
              }
         }
         
         while (i <= (int)mid - 1)   
                 b[k++] = a[i++]; 
         while (j <= (int)right)
                 b[k++] = a[j++];
         for (i=(int)left; i <= (int)right; i++) 
    	        a[i] = b[i];
    	        return c; 
         }
     
  }  