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>
const long long N = 105, P = 998244353, Q = P - 1;
long long n, m, x, y;
std::unordered_map<long long, long long> o;
struct mat {
  long long s[N][N];
} O, E, S, Z;
inline mat operator*(const mat &a, const mat &b) {
  mat c = O;
  long long i, j, k;
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++)
      for (k = 0; k < n; k++)
        c.s[i][j] = (c.s[i][j] + a.s[i][k] * b.s[k][j]) % Q;
  return c;
}
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline long long ksm(long long a, long long b, long long p) {
  long long res = 1;
  for (; b; b >>= 1, a = a * a % p)
    if (b & 1) res = res * a % p;
  return res;
}
inline mat ksm(mat a, long long b) {
  mat res = S;
  for (; b; b >>= 1, a = a * a)
    if (b & 1) res = res * a;
  return res;
}
inline long long bsgs(long long a, long long b, long long p) {
  if (m = sqrt(p) + 1, a % p < 1 && b) return -1;
  long long i, t = ksm(a, m, p);
  o.clear(), o[b] = 1;
  for (i = 1; i < m; i++) o[b = b * a % p] = i + 1;
  for (i = b = 1; i <= m; i++)
    if (o.count(b = b * t % p)) return (i * m - o[b] + 1 + p) % p;
  return -1;
}
inline long long phi(long long x) {
  long long res = x, i;
  for (i = 2; i * i <= x; i++)
    if (x % i < 1)
      for (res = res / i * (i - 1); x % i < 1; x /= i)
        ;
  if (x > 1) res = res / x * (x - 1);
  return res;
}
signed main() {
  scanf("%lld", &n);
  long long i, b;
  for (i = 0; i < n; i++)
    scanf("%lld", &b), E.s[i][i] = Z.s[i + 1][i] = 1,
                       Z.s[n - i - 1][n - 1] = b % Q;
  scanf("%lld%lld", &x, &y);
  S.s[0][n - 1] = 1, S = ksm(Z, x - n), x = bsgs(3ll, y, P),
             b = gcd(S.s[0][n - 1], Q);
  if (x % b > 0)
    puts("-1");
  else
    printf("%lld\n", ksm(3ll,
                         ksm(S.s[0][n - 1] / b, phi(Q / b) - 1, Q / b) *
                             (x / b) % (Q / b),
                         P));
  return 0;
}