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.
import java.util.*;

public class B {

	static String s, t;
	static long what = 0;
	static long[] ar = new long[2000006];
	static long[] ar2 = new long[2000006];

	public static boolean rec(long mid, long vv) {
		for (long i = 0; i < 26; i++) {
			long m = ar[(int) i];
			long mm = ar2[(int) i] * mid;
			if (mm > m) {
				vv -= (mm - m);
			}
		}
		if (vv >= 0)
			return true;
		else
			return false;
	}

	public static void main(String[] args) {
		Scanner tt = new Scanner(System.in);
		s = tt.next();
		t = tt.next();
		for (char i : s.toCharArray()) {
			if (i == '?')
				what = what + 1;
			else
				ar[i - 'a']++;
		}
		for (char ii : t.toCharArray()) {
			ar2[ii - 'a']++;
		}
		long lo = 0, hi = s.length();
		long r = 0;
		while (lo <= hi) {
			long mid = (lo + hi) / 2;
			if (rec(mid, what) == true) {
				r = mid;
				lo = mid + 1;
			} else
				hi = mid - 1;
		}

		// System.out.prlongln(r);

		char[] chh = new char[s.length()];

		long cnt = 0;

		for (char j : s.toCharArray()) {
			if (j == '?') {

				//System.out.prlongln("Al helal "+j);
				long ff = 0;
				for (long i = 0; i < 26; i++) {
					long m = ar[(int) i];
					long mm = ar2[(int) i] * r;
					if (mm > m) {

						// System.out.prlongln(m + " " + mm);
						//ff++;
						ar[(int) i]++;
						chh[(int) cnt] = (char) (i + 'a');
						cnt++;
						
						ff=5;
						break;
					}

				}
				if (ff == 0) {
					
					//System.out.prlongln("  a    ");
					//chh[cnt] = 'a';
					for(long h=0;h<26;h++) {
						if(ar[(int) h]<ar2[(int) h]) {
							ar[(int) h]++;
							chh[(int) cnt]=(char) (h+'a');
							cnt++;
							ff++;
							break;
						}
					}
					//cnt++;
				}
				if(ff==0) {
					chh[(int) cnt]=(char)('a');cnt++;
				}

			} else {
				
				//System.out.prlongln(cnt);
				chh[(int) cnt] = j;
				cnt++;
			}
			// cnt++;
		}
		String ans = String.valueOf(chh);
		System.out.println(ans);

	}

}