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;
const long long mod7 = 998244353;
vector<vector<long long> > mul(vector<vector<long long> > &A,
                               vector<vector<long long> > &B,
                               long long mod = mod7) {
  vector<vector<long long> > res(A.size(), vector<long long>(B[0].size()));
  for (size_t i = 0; i < A.size(); ++i)
    for (size_t j = 0; j < B[0].size(); ++j)
      for (size_t k = 0; k < B.size(); ++k)
        res[i][j] = (res[i][j] + A[i][k] * B[k][j]) % mod;
  return res;
}
vector<vector<long long> > pow(vector<vector<long long> > A, long long n,
                               long long mod = mod7) {
  vector<vector<long long> > res(A.size(), vector<long long>(A.size()));
  for (size_t i = 0; i < A.size(); ++i) res[i][i] = 1;
  for (; n; A = mul(A, A, mod), n >>= 1)
    if (n & 1) res = mul(res, A, mod);
  return res;
}
inline long long gcd(long long a, long long b) {
  while (a && b) a > b ? a %= b : b %= a;
  return a + b;
}
inline void exgcd(long long a, long long b, long long &x, long long &y) {
  for (x = 1, y = 0; b;) {
    long long t = a / b;
    swap(a -= t * b, b);
    swap(x -= t * y, y);
  }
}
inline long long mmul(long long a, long long b, long long mod = mod7) {
  a *= b;
  return a >= mod ? a % mod : a;
}
inline long long minv(long long a, long long mod = mod7) {
  if (gcd(a, mod) != 1) return -1;
  long long x, y;
  exgcd(a, mod, x, y);
  return x < 0 ? x + mod : x;
}
inline long long mdiv(long long a, long long b, long long mod = mod7) {
  if (a == 0) return 0;
  long long g = gcd(b, mod);
  if (a % g) return -1;
  a /= g, b /= g, mod /= g;
  return mmul(a, minv(b, mod), mod);
}
inline long long mpow(long long a, long long n, long long mod = mod7) {
  long long res = 1;
  for (; n; n >>= 1, a = mmul(a, a, mod))
    if (n & 1) res = mmul(res, a, mod);
  return res;
}
inline long long mlog(long long a, long long b, long long mod = mod7) {
  int s = sqrt(mod) + 1;
  long long t = 1;
  unordered_map<long long, long long> m;
  for (int i = 0; i < s; ++i, t = mmul(t, a, mod)) m[t] = i;
  t = minv(t, mod);
  for (int i = 0; i < s; ++i, b = mmul(t, b, mod))
    if (m.count(b)) return m[b] + i * s;
  throw;
  return -1;
}
long long find_generator(long long order, long long p) {
  vector<int> v;
  long long o = order;
  for (int i = 2; 1ll * i * i <= o; ++i)
    if (o % i == 0) {
      v.push_back(order / i);
      while (o % i == 0) o /= i;
    }
  if (o > 1) v.push_back(order / o);
  for (long long i = 2, t;; ++i) {
    for (int j : v)
      if ((t = mpow(i, j, p)) == 1ll) break;
    if (t != 1ll) return i;
  }
}
inline long long find_order(long long n) {
  long long res = n;
  for (long long i = 2; i * i <= n; ++i)
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      res /= i;
      res *= (i - 1);
    }
  if (n > 1) res /= n, res *= (n - 1);
  return res;
}
inline long long mroot(long long exp, long long b, long long mod = mod7) {
  long long o = find_order(mod);
  long long g = find_generator(o, mod);
  long long o2 = mlog(g, b, mod);
  long long o1 = mdiv(o2, exp, o);
  if (o1 == -1) return -1;
  return mpow(g, o1, mod);
}
long long N, M, K;
long long arr[105];
void init() {
  cin >> K;
  for (int i = 0; i < K; ++i) cin >> arr[i];
  cin >> N >> M;
}
void sol() {
  vector<vector<long long> > res(K, vector<long long>(K));
  for (int i = 0; i < K; ++i) res[i][0] = arr[i];
  for (int i = 1; i < K; ++i) res[i - 1][i] = 1;
  res = pow(res, N - K, mod7 - 1);
  N = res[0][0];
  cout << mroot(N, M, mod7) << endl;
}
int main() {
  init();
  sol();
}