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>
template <class T>
inline T read() {
  T r = 0, f = 0;
  char c;
  while (!isdigit(c = getchar())) f |= (c == '-');
  while (isdigit(c)) r = (r << 1) + (r << 3) + (c ^ 48), c = getchar();
  return f ? -r : r;
}
inline long long qpow(long long a, int b) {
  long long ans = 1;
  for (; b; b >>= 1) {
    if (b & 1) (ans *= a) %= 998244353;
    (a *= a) %= 998244353;
  }
  return ans;
}
int exgcd(int a, int b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  int d = exgcd(b, a % b, y, x);
  y -= x * (a / b);
  return d;
}
int K, n, m;
long long f[111], a[111][111];
long long b[111], c[111][111];
inline void Mul() {
  for (int i = 0; i < K; i++)
    for (int j = 0; j < K; j++) (b[i] += a[i][j] * f[j]) %= 998244352;
  for (int i = 0; i < K; i++) f[i] = b[i], b[i] = 0;
}
inline void Mulself() {
  for (int i = 0; i < K; i++)
    for (int j = 0; j < K; j++)
      for (int k = 0; k < K; k++) (c[i][j] += a[i][k] * a[k][j]) %= 998244352;
  for (int i = 0; i < K; i++)
    for (int j = 0; j < K; j++) a[i][j] = c[i][j], c[i][j] = 0;
}
std::map<int, int> h;
inline int BSGS(int x) {
  int t = sqrt(998244353);
  long long mul = x;
  for (int i = 0; i < t; i++) h[mul] = i, (mul *= 3) %= 998244353;
  mul = qpow(3, t);
  long long now = mul;
  for (int i = 1; i <= t + 1; i++)
    if (h.count(now))
      return i * t - h[now];
    else
      (now *= mul) %= 998244353;
  return -1;
}
int main() {
  K = read<int>();
  for (int i = 0; i < K - 1; i++) a[i][i + 1] = 1;
  for (int i = 0; i < K; i++) a[K - 1][K - i - 1] = read<int>();
  f[K - 1] = 1;
  n = read<int>();
  for (n--; n; n >>= 1) {
    if (n & 1) Mul();
    Mulself();
  }
  long long x = BSGS(read<int>());
  long long y = f[0], inv = 0, Y = 0;
  int d = exgcd(y, 998244352, inv, Y);
  if (x % d) return puts("-1"), 0;
  inv = (inv % 998244352 + 998244352) % 998244352;
  printf("%lld\n", qpow(3, x / d * inv % 998244352));
  return 0;
}