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;
int K;
vector<long long> b;
vector<vector<long long> > mul(vector<vector<long long> > A,
                               vector<vector<long long> > B) {
  vector<vector<long long> > C(K, vector<long long>(K));
  for (int i = 0; i < (int)K; i++)
    for (int j = 0; j < (int)K; j++)
      for (int k = 0; k < (int)K; k++) {
        C[i][j] += (A[i][k] * B[k][j]) % (998244353 - 1);
        C[i][j] %= (998244353 - 1);
      }
  return C;
}
vector<vector<long long> > pow(vector<vector<long long> > A, int p) {
  if (p == 1) return A;
  if (p % 2) return mul(A, pow(A, p - 1));
  vector<vector<long long> > X = pow(A, p / 2);
  return mul(X, X);
}
vector<long long> mult(vector<vector<long long> > A, vector<long long> v) {
  vector<long long> ret(K);
  for (int i = 0; i < (int)K; i++)
    for (int j = 0; j < (int)K; j++) {
      ret[i] += (A[i][j] * v[j]) % (998244353 - 1);
      ret[i] %= (998244353 - 1);
    }
  return ret;
}
long long get(int N) {
  vector<long long> F1(K);
  F1[0] = 1;
  vector<vector<long long> > T(K, vector<long long>(K));
  for (int r = 0; r < (int)K - 1; r++) T[r + 1][r] = 1;
  for (int c = 0; c < (int)K; c++) T[0][c] = b[c];
  if (N == 1) return 1;
  T = pow(T, N - K);
  return mult(T, F1)[0];
}
long long exp(long long a, long long p) {
  if (!a) return 0;
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * a) % 998244353;
    a = (a * a) % 998244353;
    p >>= 1;
  }
  return ans;
}
long long discrete_log(long long a, long long b) {
  long long n = sqrt(998244353) + 1;
  unordered_map<long long, long long> f1;
  long long an = 1, anp = 1;
  for (int i = 0; i < (int)n; i++) an = (an * a) % 998244353;
  for (int p = 0; p <= n; p++) {
    f1[anp] = p;
    anp = (anp * an) % 998244353;
  }
  long long aq = 1;
  for (int q = 0; q <= n; q++) {
    long long f2 = (b * aq) % 998244353;
    if (f1.find(f2) != f1.end())
      return (n * f1[f2] - q + 998244353) % 998244353;
    aq = (aq * a) % 998244353;
  }
  return -1;
}
long long discrete_root(long long k, long long a) {
  long long gk = exp(3, k);
  long long y = discrete_log(gk, a);
  if (y == -1) return -1;
  return exp(3, y);
}
int main() {
  cin >> K;
  b.resize(K);
  for (int i = 0; i < (int)K; i++) cin >> b[i];
  long long n, m;
  cin >> n >> m;
  long long ret = get(n);
  long long ans = discrete_root(ret, m);
  cout << ans << "\n";
}