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:
/*
     * Code Author: Anamika Modi
     * DA-IICT
     */
    import java.io.*;
import java.math.BigInteger;
import java.math.RoundingMode;
import java.text.DecimalFormat;
import java.util.*;
     
     
    public class code1 {
     
    	static double eps=(double)1e-6;
    	static long mod=(int)1e9+7;
    	public static void main(String args[]){
    		InputReader in = new InputReader(System.in);
    		OutputStream outputStream = System.out;
    		PrintWriter out = new PrintWriter(outputStream);
    		//----------My Code----------
            int n=in.nextInt();
            int b[]=new int[n];
            int a[][]=new int[n][n];
            for(int i=0;i<n;i++){
            	for(int j=0;j<n;j++){
            		a[i][j]=in.nextInt();
            	}
            }
            int f=0;
            for(int i=0;i<n;i++){
            	int k=maxRepeating(a[i],n,f);
            	if(k==n)
            		f=1;
            	            	b[i]=k;
            	
            }
    		for(int i=0;i<n;i++){
    			System.out.print(b[i]+" ");
    		}
    		
    		
    		//---------------The End------------------
    	}
    	static int maxRepeating(int arr[], int n,int f)
        {
    		int c[]=new int[n];
            // Iterate though input array, for every element
            // arr[i], increment arr[arr[i]%k] by k
            for (int i = 0; i< n; i++)
                c[arr[i]] +=1;
     
            // Find index of the maximum repeating element
            int flag=0;
            int max = Integer.MIN_VALUE, result = 0;
            for (int i = 1; i < n; i++)
            {
                if (c[i] > max)
                {
                    max = c[i];
                    result = i;
                }
                if(c[i]==max){
                	flag++;
                	
            }
            }
                if(flag==(n-1)&&f==0)
                return n;
                else if(flag==(n-1)&&f==1)
                return (n-1);
                else 
                return result;
        }
    	static long modulo(long a,long b,long c) {
    	    long x=1;
    	    	    long y=a;
    	    	    while(b > 0){
    	    	        if(b%2 == 1){
    	    	            x=(x*y)%c;
    	    	        }
    	    	        y = (y*y)%c; // squaring the base
    	    	        b /= 2;
    	    	    }
    	    	    return  x%c;
    	    	}
    	static long gcd(long x, long y)
        {
            long r=0, a, b;
            a = (x > y) ? x : y; // a is greater number
            b = (x < y) ? x : y; // b is smaller number
            r = b;
            while(a % b != 0)
            {
                r = a % b;
                a = b;
                b = r;
            }
            return r;
        }
    	static class Pair implements Comparable<Pair>{
    		long x;
    		long y;
    		int i;
    		Pair(long xx,long yy){
    			x=xx;
    			y=yy;
    		}
    		@Override
    		public int compareTo(Pair o) {
    			if(this.y==o.y)
					return Long.compare((this.x),(o.x));
				else
					return Long.compare((this.y),(o.y));
    		}
    	}
    	
    	static class InputReader {
    	    public BufferedReader reader;
    	    public StringTokenizer tokenizer;
     
    	    public InputReader(InputStream inputstream) {
    	      reader = new BufferedReader(new InputStreamReader(inputstream));
    	      tokenizer = null;
    	    }
    	    
    	    public String nextLine(){
    	    	String fullLine=null;
    	    	while (tokenizer == null || !tokenizer.hasMoreTokens()) {
    	            try {
    	              fullLine=reader.readLine();
    	            } catch (IOException e) {
    	              throw new RuntimeException(e);
    	            }
    	            return fullLine;
    	          }
    	          return fullLine;
    	    }
    		public String next() {
    	      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
    	        try {
    	          tokenizer = new StringTokenizer(reader.readLine());
    	        } catch (IOException e) {
    	          throw new RuntimeException(e);
    	        }
    	      }
    	      return tokenizer.nextToken();
    	    }
    		public long nextLong() {
    		      return Long.parseLong(next());
    		    }
    	    public int nextInt() {
    	      return Integer.parseInt(next());
    	    }
    	  }
    }