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;
vector<vector<long long> > operator*(vector<vector<long long> > A,
                                     vector<vector<long long> > B) {
  int k = A.size();
  vector<vector<long long> > C(k, vector<long long>(k));
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < k; j++) {
      C[i][j] = 0;
      for (int h = 0; h < k; h++) {
        C[i][j] += (A[i][h] * B[h][j]) % (long long)998244352;
        C[i][j] %= (long long)998244352;
      }
    }
  }
  return C;
};
vector<vector<long long> > powmod(vector<vector<long long> > A, long long n) {
  int k = A.size();
  vector<vector<long long> > res(k, vector<long long>(k));
  for (int i = 0; i < k; i++)
    for (int j = 0; j < k; j++) res[i][j] = (i == j) ? 1 : 0;
  while (n) {
    if (n & 1) {
      res = A * res;
      n--;
    } else
      A = A * A, n >>= 1;
  }
  return res;
}
long long powmod(long long a, long long b, long long p) {
  long long res = 1;
  while (b)
    if (b & 1)
      res = (long long)((res * 1ll * a) % p), --b;
    else
      a = (long long)((a * 1ll * a) % p), b >>= 1;
  return res;
}
long long bsgs(long long a, long long m) {
  a %= (long long)998244353;
  m %= (long long)998244353;
  long long n = sqrt((long long)998244353 + .0) + 1;
  map<long long, long long> vals;
  for (long long p = 1; p <= n; p++)
    vals[powmod(a, p * n, (long long)998244353)] = p;
  for (long long q = 0; q < n + 1; q++) {
    long long cur =
        ((powmod(a, q, (long long)998244353)) * m) % (long long)998244353;
    if (vals.count(cur)) {
      long long ans = vals[cur] * n - q;
      return ans;
    }
  }
  return -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  int k;
  cin >> k;
  vector<long long> b(k);
  for (int i = 0; i < k; i++) cin >> b[i];
  long long n, m;
  cin >> n >> m;
  vector<vector<long long> > B(k, vector<long long>(k));
  for (int i = 0; i < k; i++)
    for (int j = 0; j < k; j++) {
      if (i == 0)
        B[i][j] = b[j];
      else
        B[i][j] = (i == j + 1) ? 1 : 0;
    }
  long long p = (powmod(B, n - k)[0][0]);
  long long a = powmod(3, p, (long long)998244353);
  long long c = bsgs(a, m);
  if (c == -1) {
    cout << -1;
    return 0;
  }
  cout << powmod(3, c, (long long)998244353) % (long long)998244353 << endl;
}