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;
long long m;
const long long mod = 998244353;
long long bin_num(long long num, long long pow) {
  if (pow == 0) {
    return 1LL;
  }
  if (pow == 1) {
    return num % mod;
  }
  if (pow == 2) {
    return ((num * num) % mod);
  }
  long long ans = bin_num(num, pow / 2);
  ans = (ans * ans) % mod;
  if (pow % 2 == 1) {
    ans = (ans * num) % mod;
  }
  return ans;
}
long long shanks(long long a, long long b) {
  map<long long, long long> values;
  long long n = (long long)sqrt(mod) + 1;
  for (long long i = n; i >= 1; i--) {
    values[bin_num(a, i * n)] = i;
  }
  for (long long i = 0; i <= n; i++) {
    long long cur = (bin_num(a, i) * b) % mod;
    if (values.count(cur) > 0) {
      long long ans = values[cur] * n - i;
      if (ans < mod) {
        return ans;
      }
    }
  }
  return -1;
}
vector<vector<long long>> mul(const vector<vector<long long>>& one,
                              const vector<vector<long long>>& second) {
  vector<vector<long long>> x(m, vector<long long>(m, 0));
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < m; j++) {
      for (int k = 0; k < m; k++) {
        x[i][j] =
            (x[i][j] + ((one[i][k] * second[k][j]) % (mod - 1))) % (mod - 1);
      }
    }
  }
  return x;
}
vector<vector<long long>> bin_(const vector<vector<long long>>& m,
                               long long pow) {
  if (pow == 1) {
    return m;
  }
  if (pow == 2) {
    return mul(m, m);
  }
  vector<vector<long long>> t_ = bin_(m, pow / 2);
  t_ = mul(t_, t_);
  if (pow % 2 == 1) {
    t_ = mul(t_, m);
  }
  return t_;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> m;
  vector<long long> b(m);
  for (int i = 0; i < m; i++) {
    cin >> b[i];
  }
  long long n, k;
  cin >> n >> k;
  vector<vector<long long>> t_(m, vector<long long>(m, 0));
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < m; j++) {
      if (j == 0) {
        t_[i][j] = b[i];
      } else if (j == i + 1) {
        t_[i][j] = 1;
      }
    }
  }
  long long pow_ = bin_(t_, n - m)[0][0];
  long long new_os = bin_num(3, pow_);
  long long ans = shanks(new_os, k);
  if (ans == -1) {
    cout << -1 << endl;
    return 0;
  }
  ans = bin_num(3, ans);
  cout << ans << endl;
  return 0;
}