Problem: Lunar New Year is approaching, and Bob received a gift from his friend recently — a recursive sequence! He loves this sequence very much and wants to play with it.

Let f_1, f_2, …, f_i, … be an infinite sequence of positive integers. Bob knows that for i > k, f_i can be obtained by the following recursive equation:

$$$f_i = \left(f_{i - 1} ^ {b_1} ⋅ f_{i - 2} ^ {b_2} ⋅ ⋅⋅⋅ ⋅ f_{i - k} ^ {b_k}\right) mod p,$$$

which in short is

$$$f_i = \left(∏_{j = 1}^{k} f_{i - j}^{b_j}\right) mod p,$$$

where p = 998 244 353 (a widely-used prime), b_1, b_2, …, b_k are known integer constants, and x mod y denotes the remainder of x divided by y.

Bob lost the values of f_1, f_2, …, f_k, which is extremely troublesome – these are the basis of the sequence! Luckily, Bob remembers the first k - 1 elements of the sequence: f_1 = f_2 = … = f_{k - 1} = 1 and the n-th element: f_n = m. Please find any possible value of f_k. If no solution exists, just tell Bob that it is impossible to recover his favorite sequence, regardless of Bob's sadness.

Input

The first line contains a positive integer k (1 ≤ k ≤ 100), denoting the length of the sequence b_1, b_2, …, b_k.

The second line contains k positive integers b_1, b_2, …, b_k (1 ≤ b_i < p).

The third line contains two positive integers n and m (k < n ≤ 10^9, 1 ≤ m < p), which implies f_n = m.

Output

Output a possible value of f_k, where f_k is a positive integer satisfying 1 ≤ f_k < p. If there are multiple answers, print any of them. If no such f_k makes f_n = m, output -1 instead.

It is easy to show that if there are some possible values of f_k, there must be at least one satisfying 1 ≤ f_k < p.

Examples

Input


3
2 3 5
4 16


Output


4


Input


5
4 7 1 5 6
7 14187219


Output


6


Input


8
2 3 5 6 1 7 9 10
23333 1


Output


1


Input


1
2
88888 66666


Output


-1


Input


3
998244352 998244352 998244352
4 2


Output


-1


Input


10
283 463 213 777 346 201 463 283 102 999
2333333 6263423


Output


382480067

Note

In the first sample, we have f_4 = f_3^2 ⋅ f_2^3 ⋅ f_1^5. Therefore, applying f_3 = 4, we have f_4 = 16. Note that there can be multiple answers.

In the third sample, applying f_7 = 1 makes f_{23333} = 1.

In the fourth sample, no such f_1 makes f_{88888} = 66666. Therefore, we output -1 instead.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main{
    public static void main(String[] args){
        try {
            new Main().solve();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    int[]b;
    int mod=998244353;
    private void solve() throws Exception {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

        int k=in.nextInt();
        b=new int[k];
        for(int i=0;i<k;i++)b[i]=in.nextInt();
        int n=in.nextInt(),m=in.nextInt();
        int[][]ma=new int[k][k];
        for(int i=0;i<k;i++){
            ma[i][0]=b[i];
            if(i!=k-1) ma[i][i+1]=1;
        }
        int[][]x=new int[1][k];x[0][0]=1;
        int phi=mod-1;
        int a=modMul(modPow(ma,n-k,phi),x,phi)[0][0];
        int b=bsgs(m,mod);
        int[] gcd=exgcd(a,phi);
        gcd[1]=(gcd[1]%phi+phi)%phi;
        if(b%gcd[0]!=0)out.println(-1);
        else out.println(FastPowMod(3,(long)gcd[1]*(b/gcd[0]),mod));
        out.flush();
    }
    public int bsgs(int a,int mod){
        long sq=(long)Math.ceil(Math.sqrt(mod));
        HashMap<Long,Integer>map=new HashMap<>();
        long x=a;
        for(int i=1;i<=sq;i++){
            x=(x*3)%mod;
            map.put(x,i);
        }
        x=1;
        long a1=FastPowMod(3,sq,mod);
        for(int i=1;i<=sq;i++){
            x=(x*a1)%mod;
            if(map.containsKey(x))return (int)sq*i-map.get(x);
        }
        return -1;
    }
    public static int FastPowMod(int a,long b,int c){
        a=a%c;
        long ans=1;
        long a1=a;
        while(b>0){
            if((b&1)==1)
                ans=(ans*a1)%c;
            a1=(a1*a1)%c;
            b>>=1;
        }
        return (int)ans;
    }
    public static int[] exgcd(int a,int b){
        if(b==0) {
            return new int[]{a,1,0};
        }
        int[]res=exgcd(b,a%b);
        int tmp=res[1];
        res[1]=res[2];
        res[2]=tmp-(a/b)*res[2];
        return res;
    }
    public int[][] modPow(int[][]a,int b,int mod){
        int[][]ans=new int[a.length][a[0].length];
        for(int i=0;i<ans.length;i++)ans[i][i]=1;
        while(b>0){
            if((b&1)!=0)ans=modMul(ans,a,mod);
            a=modMul(a,a,mod);
            b>>=1;
        }
        return ans;
    }


    public int[][] modMul(int[][]a,int[][]b,int mod) {
        int row = a.length, column = b[0].length;
        int len = b.length;
        int[][] ans = new int[row][column];
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < len; j++) {
                if (a[i][j] == 0) continue;//稀疏矩阵优化
                for (int k = 0; k < column; k++) {
                    ans[i][k] += (long) a[i][j] * b[j][k] % mod;
                    ans[i][k] %= mod;
                }
            }
        }
        return ans;
    }
    class InputReader{
        StreamTokenizer tokenizer;
        public InputReader(InputStream stream){
            tokenizer=new StreamTokenizer(new BufferedReader(new InputStreamReader(stream)));
            tokenizer.ordinaryChars(33,126);
            tokenizer.wordChars(33,126);
        }
        public String next() throws IOException {
            tokenizer.nextToken();
            return tokenizer.sval;
        }
        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }
        public boolean hasNext() throws IOException {
            int res=tokenizer.nextToken();
            tokenizer.pushBack();
            return res!=tokenizer.TT_EOF;
        }
    }
}