/*
 * ACM ICPC - CERC 2011
 *
 * Sample solution: Vigenere Cipher Analysis (analyse)
 * Author: Martin Kacer
 */

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.HashSet;
import java.util.Set;
import java.util.StringTokenizer;

public class analyse {
	StringTokenizer st = new StringTokenizer("");
	BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
	public static void main(String[] args) throws Exception {
		analyse instance = new analyse();
		while (instance.run()) {/*repeat*/}
	}
	String nextToken() throws Exception {
		while (!st.hasMoreTokens()) st = new StringTokenizer(input.readLine());
		return st.nextToken();
	}
	int nextInt() throws Exception {
		return Integer.parseInt(nextToken());
	}
	
	String sumString(String key, CharSequence txt, int start, int end, boolean plus) {
		if (end == -1) end = txt.length();
		StringBuilder sb = new StringBuilder();
		for (int i = start; i < end; ++i) {
			int offs = key.charAt((i-start) % key.length()) - 'A' + 1;
			if (!plus) offs = -offs;
			char c = (char)(txt.charAt(i) + offs);
			sb.append((c < 'A') ? (char)(c+26) : (c > 'Z') ? (char)(c-26) : c);
		}
		return sb.toString();
	}

	int maxKeyLen;
	Set<String> posKeys1 = new HashSet<String>(100000), posKeys2 = new HashSet<String>(10000);
	
	boolean isRepetition(String key, int len) {
		for (int i = len; i < key.length(); ++i)
			if (key.charAt(i) != key.charAt(i-len)) return false;
		return true;
	}

	/** We are lazy, so instead of avoiding overlap at all, we detect it here. */
	boolean containsBoth(String txt, String s1, String s2) {
		for (int i1 = txt.indexOf(s1); i1 >=0; i1 = txt.indexOf(s1, i1+1))
			for (int i2 = txt.indexOf(s2); i2 >=0; i2 = txt.indexOf(s2, i2+1)) {
				if (i2 >= i1 + s1.length() || i1 >= i2 + s2.length()) return true;
			}
		return false;
	}
	
	boolean run() throws Exception {
		maxKeyLen = nextInt();
		if (maxKeyLen == 0) return false;
		posKeys1.clear(); posKeys2.clear();
		String s1 = nextToken(), s2 = nextToken(), msg = nextToken();
		for (int start = 0; start <= msg.length() - s1.length(); ++start) {
			String key = sumString(s1, msg, start, start+s1.length(), false);
			for (int i = 1; i <= maxKeyLen; ++i)
				if (isRepetition(key, i)) {
					String k = (i == key.length()) ? key : key.substring(0, i);
					int idx = i - start%i;
					if (idx < i) k = k.substring(idx) + k.substring(0, idx);
					posKeys1.add(k);
				}
		}
		for (int start = 0; start <= msg.length() - s2.length(); ++start) {
			String key = sumString(s2, msg, start, start+s2.length(), false);
			for (int i = 1; i <= maxKeyLen; ++i)
				if (isRepetition(key, i)) {
					String k = (i == key.length()) ? key : key.substring(0, i);
					int idx = i - start%i;
					if (idx < i) k = k.substring(idx) + k.substring(0, idx);
					//System.err.println("DBG2: " + key + "/" + i + "->" + k);
					if (posKeys1.contains(k)) posKeys2.add(k);
				}
		}
		//System.err.println("1: " + posKeys1); System.err.println("2: " + posKeys2);
		String restxt = null;
		for (String k : posKeys2) {
			String txt = sumString(k, msg, 0, -1, false);
			if (containsBoth(txt, s1, s2)) {
				if (restxt != null && !restxt.equals(txt)) {
					System.out.println("ambiguous");
					return true;
				} else
					restxt = txt;
			}
		}
		System.out.println((restxt == null) ? "impossible" : restxt);
		return true;
	}
}