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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long Readin() {
  long long K = 0, F = 1;
  char C = ' ';
  while (C < '0' or C > '9') F = C == '-' ? -1 : 1, C = getchar();
  while (C <= '9' and C >= '0')
    K = (K << 1) + (K << 3) + C - '0', C = getchar();
  return F * K;
}
const int Mod = 998244352;
const int MaxK = 100 + 5;
int N;
inline int Pow(int A, int B) {
  int Mod = 998244353;
  register int Ans = 1;
  while (B) {
    if (B & 1) Ans = 1ll * Ans * A % Mod;
    A = 1ll * A * A % Mod;
    B >>= 1;
  }
  return Ans;
}
struct Matrix {
  long long A[MaxK][MaxK];
  Matrix() { memset(A, 0, sizeof A); }
};
inline Matrix operator*(Matrix A, Matrix B) {
  register Matrix Ans;
  for (register int i = 1; i <= N; ++i)
    for (register int j = 1; j <= N; ++j)
      for (register int k = 1; k <= N; ++k)
        Ans.A[i][j] = (Ans.A[i][j] + (1ll * A.A[i][k] * B.A[k][j]) % Mod) % Mod;
  return Ans;
}
inline Matrix operator^(Matrix S, long long T) {
  register Matrix Ans;
  for (register int i = 1; i <= N; ++i) Ans.A[i][i] = 1;
  while (T) {
    if (T & 1) Ans = Ans * S;
    S = S * S;
    T >>= 1;
  }
  return Ans;
}
Matrix Give, Anss;
unordered_map<int, int> Map;
int A, B, C, P, G, Ans;
inline int Mul(int A, int B, int P) { return 1ll * A * B % P; }
int Gcd(int A, int B) { return B ? Gcd(B, A % B) : A; }
void ExGcd(int A, int B, int &X, int &Y) {
  if (B == 0) {
    X = 1;
    Y = 0;
    return;
  }
  ExGcd(B, A % B, Y, X);
  Y -= A / B * X;
}
inline int ExBsgs() {
  Map.clear();
  if (B == 1) return 0;
  register int Ans = 0;
  C = 1;
  while ((G = Gcd(A, P)) != 1) {
    if (B % G) return -1;
    ++Ans;
    B /= G;
    P /= G;
    C = Mul(C, A / G, P);
    if (C == B) return Ans;
  }
  register int Sqr = sqrt(P) + 1, T = 1, T2;
  Map.clear();
  for (register int i = 0; i < Sqr; ++i) {
    Map[Mul(T, B, P)] = i;
    T = Mul(T, A, P);
  }
  T2 = Mul(T, C, P);
  for (register int i = 1; i <= Sqr; ++i, T2 = Mul(T2, T, P))
    if (Map.find(T2) != Map.end()) return i * Sqr - Map[T2] + Ans;
  return -1;
}
int main() {
  N = Readin();
  for (register int i = 0; i < N; ++i) Give.A[N - i][N] = Readin();
  for (register int i = 2; i <= N; ++i) Give.A[i][i - 1] = 1;
  int K = Readin(), M = Readin();
  Anss.A[1][N] = 1;
  Give = Give ^ (K - N);
  Anss = Anss * Give;
  int Q = Anss.A[1][N];
  int G = 3;
  A = G;
  B = M;
  P = Mod + 1;
  int J = ExBsgs();
  P = Mod;
  register int Gc = Gcd(P, Q);
  if (J % Gc) return not printf("-1\n");
  int X, Y;
  ExGcd(Q, P, X, Y);
  X = (X % P + P) % P;
  Ans = 1ll * X * (J / Gc) % P;
  return not printf("%d\n", Pow(G, Ans));
}