Developers often face with regular expression patterns. A pattern is usually defined as a string consisting of characters and metacharacters that sets the rules for your search. These patterns are most often used to check whether a particular string meets the certain rules.

In this task, a pattern will be a string consisting of small English letters and question marks ('?'). The question mark in the pattern is a metacharacter that denotes an arbitrary small letter of the English alphabet. We will assume that a string matches the pattern if we can transform the string into the pattern by replacing the question marks by the appropriate characters. For example, string aba matches patterns: ???, ??a, a?a, aba.

Programmers that work for the R1 company love puzzling each other (and themselves) with riddles. One of them is as follows: you are given n patterns of the same length, you need to find a pattern that contains as few question marks as possible, and intersects with each of the given patterns. Two patterns intersect if there is a string that matches both the first and the second pattern. Can you solve this riddle?

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of patterns. Next n lines contain the patterns.

It is guaranteed that the patterns can only consist of small English letters and symbols '?'. All patterns are non-empty and have the same length. The total length of all the patterns does not exceed 105 characters.

Output

In a single line print the answer to the problem — the pattern with the minimal number of signs '?', which intersects with each of the given ones. If there are several answers, print any of them.

Examples

Input

2
?ab
??b


Output

xab


Input

2
a
b


Output

?


Input

1
?a?b


Output

cacb

Note

Consider the first example. Pattern xab intersects with each of the given patterns. Pattern ??? also intersects with each of the given patterns, but it contains more question signs, hence it is not an optimal answer. Clearly, xab is the optimal answer, because it doesn't contain any question sign. There are a lot of other optimal answers, for example: aab, bab, cab, dab and so on.
import java.io.*;
import java.math.BigInteger;
import java.util.*;
 
 
 
 public class Main {
	public static InputReader in;
    public static PrintWriter pw;
 
 
 
	public static void main(String args[]) {
		new Thread(null, new Runnable() {
            public void run() {
                try{
                    solve();
                }
                catch(Exception e){
                    e.printStackTrace();
                }
            }
        }, "1", 1 << 26).start();
    }
	static ArrayList<Integer> g[];
	static ArrayList<Integer> h[];
	static boolean visited[];
	static boolean visit[];
	static int Parent[];
	static int degree[];
	static int edjes=0;
	static int start=-1;
	static int end=-1;
	static int Total=0;
	static int time1[];
	static int time2[];
	static int glob=0;
	static int ans[];
	static boolean vis2[][];
	static long sum1=0;
	static long val[];
    static ArrayList<Integer> levels[];
    static int max=0;
    static int lev=1;
    static ArrayList<Integer> nodes[];
    static ArrayList<Integer> values[];
    static int depth[];
    static boolean found=false;
    static boolean inStack[];
    public static void solve(){
		
	
		 in = new InputReader(System.in);
		 pw = new PrintWriter(System.out);
         int n=in.nextInt();
         int q[]=new int[100004];
         String s1=in.readString();
         TreeSet<Character> ss[]=new TreeSet[s1.length()];
         for(int i=0;i<s1.length();i++)
        	 ss[i]=new TreeSet();
         for(int j=0;j<s1.length();j++)
         {
        	 if(s1.charAt(j)=='?')
        		 q[j]++;
        	 else
        		 ss[j].add(s1.charAt(j));
         }
         for(int i=1;i<n;i++)
         {
        	 String s=in.readString();
             for(int j=0;j<s.length();j++)
             {
            	 if(s.charAt(j)=='?')
            		 q[j]++;
            	 else
            	 {
            		 ss[j].add(s.charAt(j));
            	 }
             }
         }
         //System.out.println(ss[1]);
         for(int i=0;i<s1.length();i++)
         {
        	 if(ss[i].size()>1)
        	 {
        		 System.out.print("?");
        	 }
        	 else if(q[i]==n)
        	 {
        		 System.out.print("p");
        	 }
        	 else 
        		 System.out.print(ss[i].first());
         }
        }
    
    public static int countNumber(int n)
    {
        int result = 0;
     
        // Pushing 1 to 9 because all number from 1
        // to 9 have this propert.
        for (int i = 1; i <= 9; i++)
        {
            Stack<Integer> s=new Stack();
            if (i <= n)
            {
                s.push(i);
                result++;
            }
     
            // take a number from stack and add
            // a digit smaller than last digit
            // of it.
            while (!s.empty())
            {
                int tp = s.peek();
                s.pop();
                for (int j = tp%10; j <= 9; j++)
                {
                    int x = tp*10 + j;
                 //   System.out.println(x);
                    if (x <= n)
                    {
                        s.push(x);
                        result++;
                    }
                }
              //  System.out.println();
            }
        }
     
        return result;
    }
/*public static void dfs2(int curr,int lev)
{
	time1[curr]=glob++;
	
	depth[curr]=lev;
	visited[curr]=true;
	nodes[lev].add(time1[curr]);
	values[lev].add(a[curr]);
	for(int x:g[curr])
	{
		if(!visited[x])
		{
			dfs2(x,lev+1);
		}
	}
time2[curr]=glob++;
}
 
 
 
public static void bfs1(int curr)
{
Queue<Integer> Q=new LinkedList<Integer>();
Q.add(1);
visit[1]=true;
int lev=2;
levels[1].add(1);
while(!Q.isEmpty())
{
	int cnt=Q.size();
	while(cnt>0)
	{
		int p=Q.poll();
		visit[p]=true;
		for(int x:g[p])
		{
			if(!visit[x])
			{
				Q.add(x);
				levels[lev].add(x);
			}
		}
		cnt--;
	}
	max=Math.max(lev, max);
	lev++;
}
}
 
 
	
 	private static long fact(int len) {
	long ans=1;
	for(int i=1;i<=len;i++)
		ans*=i;
	return ans;
}
 
 
 
	
	// SubTree Checker DFS (TO CHECK IF X IS IN A SUBTREE OF Y)
	public static void dfs3(int curr)
	{
		visited[curr]=true;
		time1[curr]=glob++;
		for(int x:g[curr])
		{
		if(!visited[x])
			dfs3(x);
		}
		time2[curr]=glob++;
	}
    
	
	
 
    /*NO OF EDJES IN WHOLE CONNECTED COMPONENTS
  
	public static void dfs1(int curr)
	{
		visited[curr]=true;
		for(int next:g[curr])
		{
			edje++;
			if(!visited[next])
				dfs1(next);
		}
	}
	/*SUM OF ALL SUBTREE NODE'S VLAUES DFS
	public static void dfs(int curr,int prev)
		{
		val[curr]=1;
			for(int x:g[curr])
			{
				if(x!=prev)
				{
		            dfs(x,curr);
		            val[curr]+=val[x];
					
			    }
			}
		}*/
	
		public static long power(long a,long b)
		{
			long result=1;
			while(b>0)
			{
				if(b%2==1)
					result*=a;
				a=a*a;
				b/=2;
			}
			return result;
		}
				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;
	}
	static class Pair implements Comparable<Pair>{
	int ballon;
		int weight;
		Pair(int ballon,int weight){
	     	this.ballon=ballon;
			this.weight=weight;
			
		}
		@Override
		public int compareTo(Pair o) {
		    return 1;
			}
		
	}
	public static long gcd(long a, long b) {
		  if (b == 0) return a;
		  return gcd(b, a%b);
		}
	static class InputReader {
 
		private InputStream stream;
		private byte[] buf = new byte[8192];
		private int curChar, snumChars;
		private SpaceCharFilter filter;
 
		public InputReader(InputStream stream) {
			this.stream = stream;
		}
 
		public int snext() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}
 
		public   int nextInt() {
			int c = snext();
			while (isSpaceChar(c))
				c = snext();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = snext();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = snext();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public long nextLong() {
			int c = snext();
			while (isSpaceChar(c))
				c = snext();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = snext();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = snext();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public int[] nextIntArray(int n) {
			int a[] = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}
 
		public String readString() {
			int c = snext();
			while (isSpaceChar(c))
				c = snext();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = snext();
			} while (!isSpaceChar(c));
			return res.toString();
		}
 
		public boolean isSpaceChar(int c) {
			if (filter != null)
				return filter.isSpaceChar(c);
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
 
		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
		}
	}
}
//Language: Java