Problem: Princess Twilight went to Celestia and Luna's old castle to research the chest from the Elements of Harmony.

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A sequence of positive integers bi is harmony if and only if for every two elements of the sequence their greatest common divisor equals 1. According to an ancient book, the key of the chest is a harmony sequence bi which minimizes the following expression:

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You are given sequence ai, help Princess Twilight to find the key.

Input

The first line contains an integer n (1 ≤ n ≤ 100) — the number of elements of the sequences a and b. The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 30).

Output

Output the key — sequence bi that minimizes the sum described above. If there are multiple optimal sequences, you can output any of them.

Examples

Input

5
1 1 1 1 1


Output

1 1 1 1 1 

Input

5
1 6 4 2 8


Output

1 5 3 1 8 

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class LittlePonyandHarmonyChest
{
	
	
	/************************ SOLUTION STARTS HERE ***********************/
	
	static int N , arr[];
	static int memo[][];
	static int MAX = 60;
	static int primes[] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59};
	static int inv[];
	static int primeFactors[];
	static int rec(int idx, int mask) {
		if(idx == N)
			return memo[idx][mask] = 0;
		else if(memo[idx][mask] != -1)
			return memo[idx][mask];
		else {
			int min = Math.abs(1 - arr[idx]) + rec(idx + 1, mask);

			for(int i=2;i<MAX;i++)
				if((primeFactors[i] & mask) == 0) 
					min = Math.min(min,Math.abs(i - arr[idx]) + rec(idx + 1, mask | primeFactors[i]));
			
			return memo[idx][mask] = min;
		}
	}
	
	private static void solve(FastScanner s1, PrintWriter out){
		
		N = s1.nextInt();
		arr = s1.nextIntArray(N);
		inv = new int[MAX];
		for(int i=0;i<primes.length;i++)
			inv[primes[i]] = i;
		
		primeFactors = new int[MAX];
		
		for(int p : primes) 
			for(int j=p;j<MAX;j+=p)
				primeFactors[j] |= 1 << inv[p];
		
		
		memo = new int[N + 1][1 << primes.length];
		for(int a[] : memo)
			Arrays.fill(a, -1);
		
		
		int currMask = 0;
		int minSum = rec(0, 0);
		
		for(int i=0;i<N;i++) {
			if(memo[i + 1][currMask] + Math.abs(1 - arr[i]) == minSum) {
				out.print("1 ");
				minSum -= Math.abs(1 - arr[i]);
			}
			else {
				for(int j=2;j<MAX;j++) {
					if((currMask & primeFactors[j]) == 0 && memo[i + 1][currMask | primeFactors[j]] + Math.abs(j - arr[i]) == minSum) {
						currMask |= primeFactors[j];
						minSum -= Math.abs(j - arr[i]);
						out.print(j + " ");
						break;
					}
				}
			}
		}
		
		
	}
	
	
	
	/************************ SOLUTION ENDS HERE ************************/
	
	
	
	
	
	/************************ TEMPLATE STARTS HERE *********************/
	
	public static void main(String []args) throws IOException {
		FastScanner in  = new FastScanner(System.in);
		PrintWriter out = 
				new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false); 
		solve(in, out);
		in.close();
		out.close();
	}    
	
	static class FastScanner{
		BufferedReader reader;
		StringTokenizer st;
		FastScanner(InputStream stream){reader=new BufferedReader(new InputStreamReader(stream));st=null;}	
		String next()
		{while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}		    
		st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
		String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}	    	  	
		int    nextInt()   {return Integer.parseInt(next());}
		long   nextLong()  {return Long.parseLong(next());}		
		double nextDouble(){return Double.parseDouble(next());}
		char   nextChar()  {return next().charAt(0);}
		int[]  nextIntArray(int n)         {int[] a= new int[n];   int i=0;while(i<n){a[i++]=nextInt();}  return a;}
		long[] nextLongArray(int n)        {long[]a= new long[n];  int i=0;while(i<n){a[i++]=nextLong();} return a;}	
		int[]  nextIntArrayOneBased(int n) {int[] a= new int[n+1]; int i=1;while(i<=n){a[i++]=nextInt();} return a;}	    	
		long[] nextLongArrayOneBased(int n){long[]a= new long[n+1];int i=1;while(i<=n){a[i++]=nextLong();}return a;}	    	
		void   close(){try{reader.close();}catch(IOException e){e.printStackTrace();}}				
	}
	
	/************************ TEMPLATE ENDS HERE ************************/
}