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.util.*;
import java.io.*;
public class B{
	public static void main(String[] args)
	{
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		String str = fs.next();
		char[] str1 = new char[str.length()];
		char[] str2 = fs.next().toCharArray();
		long have = 0;
		int[] count = new int[26];
		List<Integer> idx = new ArrayList();
		for(int i=0;i<str1.length;i++)
		{
			str1[i] = str.charAt(i);
			char ch = str1[i];
			if(ch!='?')count[ch-'a']++;
			else
			{
				have++;
				idx.add(i);
			}
		}
		int[] wantCount = new int[26];
		for(char ch:str2)
		{
			wantCount[ch-'a']++;
		}
		long low = 0; long high = (str1.length/str2.length) + 1;
		while(low+1<high)
		{
			long mid = (low+high)/2;
			long need = 0;
			for(int i=0;i<26;i++)
			{
				long myWillHave = (long)mid*(long)wantCount[i];
				need += (long)Math.max(0,myWillHave-count[i]);
			}
			if(need<=have)
			{
				low = mid;
			}
			else high = mid;
		}
//		if(have>(int)1e5)
//		{
//			System.out.println("Here "+low);
//		}
//		out.println("Low: "+low);
		int start = 0;
		for(int i=0;i<26;i++)
		{
			if(wantCount[i]>0)
			{
				long tot = (long)wantCount[i]*(long)low;
				long more = tot - count[i];
//				if(have>(int)1e5)System.out.println("Inside: "+more);
				for(int j=0;j<more;j++)
				{
					str1[idx.get(start)] = (char)('a'+i);
					start++;
				}
			}
		}
//		if(have>(int)1e5)
//		{
//			System.out.println("Here Outside");
//		}
		for(int i=start;i<idx.size();i++)str1[idx.get(i)] = 'a';
		for(char ch:str1)
		{
			out.print(ch);
		}
		out.println();
		out.close();
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	public static int[] sort(int[] arr)
	{
		List<Integer> temp = new ArrayList();
		for(int i:arr)temp.add(i);
		Collections.sort(temp);
		int start = 0;
		for(int i:temp)arr[start++]=i;
		return arr;
	}
	public static int[] sort2(int[] arr)
	{
		List<Integer> temp = new ArrayList();
		for(int i:arr)temp.add(i);
		Collections.sort(temp,Collections.reverseOrder());
		int start = 0;
		for(int i:temp)arr[start++]=i;
		return arr;
	}
}