/* Sample solution to D - Pseudo random numbers / Mikael Goldmann
 * First count backwards to recreate the seed number 
 * The count forwards to get T:th number. Try using few columns to
 * save space.
 */
import java.io.*;
import java.util.*;


class random_mg 
{
    static BufferedReader stdin;
    
    int[][] tab;
    public static void main(String[] ss) throws IOException
    {
	Reader rdr = new InputStreamReader(System.in);
	stdin = new BufferedReader(rdr);
	int n = Integer.parseInt(stdin.readLine());
	for (int i = 1; i <= n; ++i) 
	    (new random_mg()).solve(i);	
    }
    
    void solve(int casenum)  throws IOException
    {
	int i,j;
	
	int B = Integer.parseInt(stdin.readLine());
	String[] tok =stdin.readLine().split("[ ]+");
	int L = Integer.parseInt(tok[0]);
	int T = Integer.parseInt(stdin.readLine());
	int ncol=5;
	
	int[] vals = new int[L];
	for (i=1; i <= L; ++i) 
	    vals[i-1] = Integer.parseInt(tok[i]);
	boolean ok=false;
	while(!ok) {
	    ok = true;
	    try{
		ncol *= 2;
		tab = newtab(L,ncol);
		for (i = 0; i < L; ++i)
		    tab[i][0] = vals[i];
		for(i = L-1; i > 0; --i)
		    if (! explain(tab,i, ncol, B)) {
			System.out.println("impossible");
			return;
		    }
	    }
	    catch(ArrayIndexOutOfBoundsException e) {
		ok = false;		
		// Retry, doubling number of columns
	    }
	}
	// Now tab[0] contains the seed numbers digits

	int ncol1=ncol+1; // One extra column
	
	/* Try finding values on the T:th row
	 * If we fail and truncated any row, retry using
	 * twice as many columns
	 */
	int k;	
	while (true) {
	    boolean fullrow = false; // no overflowing row yet
	    int[][] v = new int[2][ncol1];
	    for (i = 0; i < ncol; ++i)
		v[0][i] = tab[0][i];
	    for (i = ncol; i < ncol1; ++i)
		v[0][i] = -1;

	    int[] a,b=null;;
	    int tmp;	
	    ok = true;	    
	    for (k = 1; k < T; ++k) {
		a = v[1-(k&1)];
		b = v[k&1];
		for (i = 0; i < ncol1; ++i) b[i] = -1;
		
		for (i = 1, j = 0; j < ncol1 &&  i <ncol1; ++i) {
		    if (a[i] == -1) break;
		    
		    tmp = a[i-1]+a[i];
		    if (tmp < B) b[j++] = tmp;
		    else {
			b[j++] = tmp-B;
			if (j < ncol1) b[j++] = 1;
		    }
		}
		if (j == ncol1) fullrow=true;
		if (b[0] == -1) {
		    ok = false;
		    break;
		}
	    }
	    if (ok) {
		System.out.println(b[0]);
		return;
	    }
	    else if(!fullrow) {
		System.out.println("unpredictable");
		return;
	    }
	    ncol1 *=2;
	}
    }
    
    int[][] newtab(int r, int c) 
    {
	int[][] t = new int[r][c];
	int i,j;	
	for (i=0; i < r; ++i)
	    for (j = 0; j < c; ++j) 
		t[i][j] = -1;
	return t;
	
    }
    
/* explain() computes row (i-1) from row i
 * Returns false if impossible. Used to calculate seed
 * from the given sequence
 */
    boolean explain(int[][] tab, int r, int ncol, int B)
    {
	int[] s = new int[ncol];
	int i,j;
	for (i=0; i < ncol; ++i) s[i] = -1;
	i=0;
	j=1;
	int s1,t1;	
	while (i < ncol && tab[r][i] != -1) {
	    tab[r-1][j] = (B + tab[r][i] - tab[r-1][j-1]) % B;
	    s1 = tab[r-1][j] + tab[r-1][j-1];
	    ++i;
	    ++j;
	    
	    if (s1 >= B) {
		if (tab[r][i] != -1 && tab[r][i] != 1) return false;
		++i;
	    }
	}
	return true;	
	    
    }
    
}