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import java.util.Arrays;
import java.util.Scanner;

/*
 * Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.

GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.

GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?

Input
The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).

All three strings consist only of lowercase English letters.

It is possible that b and c coincide.

Output
Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.

Sample test(s)
input
aaa
a
b
output
aaa
input
pozdravstaklenidodiri
niste
dobri
output
nisteaadddiiklooprrvz
input
abbbaaccca
ab
aca
output
ababacabcc
Note
In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc.
 */
public class B {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String str = sc.nextLine();
		String a = sc.nextLine();
		String b = sc.nextLine();
		
		int[] strCount = getCharCount(str);
		int[] aCount = getCharCount(a);
		int[] bCount = getCharCount(b);
		
		int minA = getMinCount(strCount, aCount);
		if (minA > 0) {
			for (int i=0; i<strCount.length; i++) {
				if (aCount[i] > 0 && strCount[i] > 0)
					strCount[i] -= minA;
			}
		}
		
		int minB = getMinCount(strCount, bCount);
		if (minB > 0) {
			for (int i=0; i<strCount.length; i++) {
				if (bCount[i] > 0 && strCount[i] > 0) 
					strCount[i] -= minB;
			}
		}
		
		StringBuilder res = new StringBuilder();
		for (int i=0; i<minA; i++) res.append(a);
		for (int i=0; i<minB; i++) res.append(b);
		res.append(getString(strCount));
		System.out.println(res);
	}
	
	public static String getString(int[] count) {
		StringBuilder sb = new StringBuilder();
		for (int i=0; i<count.length; i++) {
			int len = count[i];
			for (int j=0; j<len; j++) {
				char ch = (char) ((int) 'a' + i);
				sb.append(ch);
			}
		}
		return sb.toString();
	}
	
	public static int getMinCount(int[] strCount, int[] count) {
		int min = Integer.MAX_VALUE;
		for (int i=0;i<26;i++) {
			if (count[i] > 0) {
				int ratio = strCount[i]/count[i];
				if (ratio < min) min = ratio;
			}
		}
		return min;
	}
	
	public static int[] getCharCount(String str) {
		int[] charCount = new int[26];
		for (char ch : str.toCharArray())
			charCount[ch-'a']++;
		return charCount;
	}
}
