Problem: You are given two strings s and t consisting of small Latin letters, string s can also contain '?' characters. 

Suitability of string s is calculated by following metric:

Any two letters can be swapped positions, these operations can be performed arbitrary number of times over any pair of positions. Among all resulting strings s, you choose the one with the largest number of non-intersecting occurrences of string t. Suitability is this number of occurrences.

You should replace all '?' characters with small Latin letters in such a way that the suitability of string s is maximal.

Input

The first line contains string s (1 ≤ |s| ≤ 106).

The second line contains string t (1 ≤ |t| ≤ 106).

Output

Print string s with '?' replaced with small Latin letters in such a way that suitability of that string is maximal.

If there are multiple strings with maximal suitability then print any of them.

Examples

Input

?aa?
ab


Output

baab


Input

??b?
za


Output

azbz


Input

abcd
abacaba


Output

abcd

Note

In the first example string "baab" can be transformed to "abab" with swaps, this one has suitability of 2. That means that string "baab" also has suitability of 2.

In the second example maximal suitability you can achieve is 1 and there are several dozens of such strings, "azbz" is just one of them.

In the third example there are no '?' characters and the suitability of the string is 0.

Solution (in JAVA):

Solution:
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;


public class q5 {
	
	
	
	
	public static void main(String[] args) throws IOException {
		
		Reader.init(System.in);
		PrintWriter out=new PrintWriter(System.out);
		StringBuffer output=new StringBuffer("");
		String s1=Reader.next();
		String s2=Reader.next();
		int[] count=new int[26];
		int qcount=0;
		int[] count2=new int[26];
		for(int i=0;i<s1.length();i++) {
			if(s1.charAt(i)=='?') {
				qcount++;
			}
			else {
				int a=s1.charAt(i)-'a';
				count[a]++;
			}
		}
		for(int i=0;i<s2.length();i++) {
			int a=s2.charAt(i)-'a';
			count2[a]++;
		}
		int[] replace=new int[26];
		char[] ans=new char[s1.length()];
		boolean end=false;
		while(qcount>0) {
			
			for(int i=0;i<26;i++) {
				if(count[i]>=count2[i]) {
					count[i]-=count2[i];
				}
				else {
					int diff=count2[i]-count[i];
					count[i]=0;
					diff=Math.min(qcount, diff);
					qcount-=diff;
					replace[i]+=diff;
				}
			}
			
		}
		int index=0;
		for(int i=0;i<s1.length();i++) {
			if(s1.charAt(i)=='?') {
				if(replace[index]>0) {
					replace[index]--;
					ans[i]=(char)(index+'a');
				}
				else {
					index++;i--;
				}
			}
			else ans[i]=s1.charAt(i);
		}
		output.append(ans);
		out.print(output.toString());
		out.flush();
	}
}



class NoD{
	int a, b;
	NoD(int aa,int bb){
		a=Math.min(aa, bb);b=Math.max(aa, bb);
	}
}







class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;
    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                     new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }
    /** get next word */
    static String nextLine() throws IOException{
    	return reader.readLine();
    }
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                   reader.readLine() );
        }
        return tokenizer.nextToken();
    }
    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
    static long nextLong() throws IOException {
        return Long.parseLong( next() );
    }
    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
}