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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class D {
	
	static int n, m, q;
	static char[] s, t;
	static int[] sOcc, tOcc;
	
	public static void main(String[] args) throws Exception
	{
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		s = sc.next().toCharArray(); n = s.length;
		t = sc.next().toCharArray(); m = t.length;
		sOcc = new int[26]; tOcc = new int[26];
		for(int i = 0; i < n; i++)
		{
			if(s[i] == '?') {q++; continue;}
			sOcc[s[i] - 'a']++;
		}
		for(int i = 0; i < m; i++) tOcc[t[i] - 'a']++;
		int lo = 0, hi = n / m, ans = -1;
		while(lo <= hi)
		{
			int mid = (lo + hi) >> 1;
			if(can(mid))
				lo = (ans = mid) + 1;
			else hi = mid - 1;
		}
		int[] fill = new int[26];
		for(int i = 0; i < 26; i++)
			fill[i] = Math.max(0, tOcc[i] * ans - sOcc[i]);
		int idx = 0;
		for(int i = 0; i < n; i++)
			if(s[i] == '?')
			{
				while(idx < 26 && fill[idx] == 0) idx++;
				if(idx == 26) s[i] = 'a';
				else {s[i] = (char) (idx + 'a'); fill[idx]--;}
			}
		for(char c : s) out.print(c);
		out.println();
		out.flush();
		out.close();
	}
	
	static boolean can(int x)
	{
		int sum = 0;
		for(int i = 0; i < 26; i++)
			sum += Math.max(0, tOcc[i] * x - sOcc[i]);
		return sum <= q;
	}
	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {br = new BufferedReader(new InputStreamReader(system));}
		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public String nextLine()throws IOException{return br.readLine();}
		public int nextInt() throws IOException {return Integer.parseInt(next());}
		public double nextDouble() throws IOException {return Double.parseDouble(next());}
		public char nextChar()throws IOException{return next().charAt(0);}
		public Long nextLong()throws IOException{return Long.parseLong(next());}
		public boolean ready() throws IOException{return br.ready();}
		public void waitForInput(){for(long i = 0; i < 3e9; i++);}
	}
}