Problem: Bob has a permutation of integers from 1 to n. Denote this permutation as p. The i-th element of p will be denoted as pi. For all pairs of distinct integers i, j between 1 and n, he wrote the number ai, j = min(pi, pj). He writes ai, i = 0 for all integer i from 1 to n.

Bob gave you all the values of ai, j that he wrote down. Your job is to reconstruct any permutation that could have generated these values. The input will be formed so that it is guaranteed that there is at least one solution that is consistent with the information given.

Input

The first line of the input will contain a single integer n (2 ≤ n ≤ 50).

The next n lines will contain the values of ai, j. The j-th number on the i-th line will represent ai, j. The i-th number on the i-th line will be 0. It's guaranteed that ai, j = aj, i and there is at least one solution consistent with the information given.

Output

Print n space separated integers, which represents a permutation that could have generated these values. If there are multiple possible solutions, print any of them.

Examples

Input

2
0 1
1 0


Output

2 1


Input

5
0 2 2 1 2
2 0 4 1 3
2 4 0 1 3
1 1 1 0 1
2 3 3 1 0


Output

2 5 4 1 3

Note

In the first case, the answer can be {1, 2} or {2, 1}.

In the second case, another possible answer is {2, 4, 5, 1, 3}.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.StringTokenizer;


public class Codeforces2 {
	public static void main(String[] args)
	{
		Codeforces2 cm=new Codeforces2();
fastscanner in=cm.new fastscanner();
int n=in.nextInt();
int a[]=new int[n];

int m[][]=new int[n][n];
for(int i=0;i<n;i++)
{
	int ct[]=new int[n];
	
	
	for(int j=0;j<n;j++)
	{
		m[i][j]=in.nextInt();
		if(m[i][j]==n-1)
		{
			a[i]=n-1;
			a[j]=n;
			//break;
		}
		ct[m[i][j]]++;
		if(ct[m[i][j]]==2)
			a[i]=m[i][j];
	
	}
}
/*
for(int i=0;i<n;i++)
{
	int ct2[]=new int[n];
	int f=1;
	for(int j=0;j<n;j++)
	{
		if(m[j][i])
		ct2[m[j][i]]++;
		if(ct2[m[j][i]]==2)
			a[i]=m[j][i];
		else if()
	}
}*/
for(int i=0;i<n;i++)
	System.out.print(a[i]+" ");


	}
	
	 class fastscanner
	{
		BufferedReader br;
		StringTokenizer s=null;
		  public fastscanner() {
			// TODO Auto-generated constructor stub
		br=new BufferedReader(new InputStreamReader(System.in));		
		}
		String next()
	{
		while(s==null||!s.hasMoreElements())
		{
			
			try
			{
				s=new StringTokenizer(br.readLine());
			}catch(IOException e)
			{
				e.printStackTrace();
			}
	}
		return s.nextToken();
		}
		String nextLine(){
			String s="";
			try {
				s= br.readLine();
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
			return s;
			
		}
		int nextInt()
		{
			return Integer.parseInt(next());
		}
		long nextLong()
		{
			return Long.parseLong(next());
		}
		double nextDouble()
		{
			return Double.parseDouble(next());
		}
	
}}