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>> identity(int n) {
  vector<vector<long long>> I(n, vector<long long>(n));
  for (int i = 0; i < n; i++) I[i][i] = 1;
  return I;
}
vector<vector<long long>> modmul(const vector<vector<long long>>& A,
                                 const vector<vector<long long>>& B,
                                 long long mod) {
  int n = A.size();
  vector<vector<long long>> C(n, vector<long long>(n));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        C[i][j] += A[i][k] * B[k][j] % mod;
        C[i][j] %= mod;
      }
    }
  }
  return C;
}
vector<vector<long long>> modpow(vector<vector<long long>> A, long long b,
                                 long long mod) {
  vector<vector<long long>> result = identity(A.size());
  for (; b > 0; b /= 2) {
    if (b % 2) result = modmul(result, A, mod);
    A = modmul(A, A, mod);
  }
  return result;
}
long long modpow(long long a, long long b, long long mod) {
  long long result = 1;
  for (; b > 0; b /= 2) {
    if (b % 2) result = result * a % mod;
    a = a * a % mod;
  }
  return result;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
vector<long long> factor(long long n) {
  vector<long long> result;
  for (long long k = 2; k * k <= n; k++) {
    if (n % k == 0) {
      result.push_back(k);
      while (n % k == 0) n /= k;
    }
  }
  if (n > 1) result.push_back(n);
  return result;
}
long long generator(long long p) {
  long long phi = p - 1;
  vector<long long> prime_divisors = factor(phi);
  for (long long g = 2; g < p; g++) {
    bool ok = true;
    for (long long q : prime_divisors) {
      if (modpow(g, phi / q, p) == 1) {
        ok = false;
        break;
      }
    }
    if (ok) return g;
  }
  return -1;
}
long long discrete_log(long long a, long long b, long long mod) {
  long long s = (long long)sqrt(mod);
  map<long long, long long> bs;
  for (long long k = 0, x = b; k <= s; k++, x = x * a % mod) bs[x] = k;
  long long gs = modpow(a, s, mod);
  for (long long k = s, x = gs; k <= mod; k += s, x = x * gs % mod)
    if (bs.count(x)) return k - bs[x];
  return -1;
}
long long canonical(long long a, long long mod) {
  a %= mod;
  if (a < 0) a += mod;
  return a;
}
long long solve(long long a, long long b, long long mod) {
  a = canonical(a, mod);
  b = canonical(b, mod);
  if (b == 0) return 0;
  long long d = gcd(mod, a);
  if (b % d) return -1;
  long long x0 = 1, y0 = 0, z0 = mod, x1 = 0, y1 = 1, z1 = a;
  while (z1 > d) {
    long long q = z0 / z1;
    x0 -= q * x1, y0 -= q * y1;
    z0 -= q * z1;
    swap(x0, x1);
    swap(y0, y1);
    swap(z0, z1);
  }
  return canonical(b / d * y1, mod);
}
int main() {
  ios_base::sync_with_stdio(false);
  const long long p = 998244353, phi = p - 1;
  long long k, n, m;
  cin >> k;
  vector<vector<long long>> A(k, vector<long long>(k));
  for (int i = 0; i < k - 1; i++) A[i][i + 1] = 1;
  for (int i = 0; i < k; i++) cin >> A[k - 1][k - 1 - i];
  cin >> n >> m;
  A = modpow(A, n - k, phi);
  long long c = A[k - 1][k - 1];
  long long g = generator(p);
  long long h_n = discrete_log(g, m, p);
  long long h_k = solve(c, h_n, phi);
  if (h_k == -1)
    cout << -1 << endl;
  else
    cout << modpow(g, h_k, p) << endl;
}