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;
using ii = pair<int, int>;
using ll = long long;
template <typename T>
ostream& operator<<(ostream& os, vector<T> v) {
  os << "[ ";
  for (auto e : v) os << e << " ";
  return os << "]";
}
template <typename T, typename V>
ostream& operator<<(ostream& os, set<T, V> v) {
  os << "{ ";
  for (auto e : v) os << e << " ";
  return os << "}";
}
template <typename T, typename V, typename S>
ostream& operator<<(ostream& os, map<T, V, S> v) {
  os << "{ ";
  for (auto e : v) os << e << " ";
  return os << "}";
}
template <typename T, typename V>
ostream& operator<<(ostream& os, pair<T, V> x) {
  return os << "( " << x.first << ", " << x.second << " )";
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& e : v) is >> e;
  return is;
}
template <typename T, typename V>
istream& operator>>(istream& is, pair<T, V>& x) {
  return is >> x.first >> x.second;
}
template <typename T>
T divfloor(T a, T b) {
  assert(b != 0);
  if (a >= 0 && b > 0 || a <= 0 && b < 0) return a / b;
  return (a % b == 0) ? a / b : a / b - 1;
}
template <typename T>
T divceil(T a, T b) {
  assert(b != 0);
  if (a >= 0 && b > 0 || a <= 0 && b < 0)
    return (a % b == 0) ? a / b : a / b + 1;
  return a / b;
}
void solve();
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
}
vector<vector<ll>> id(int n) {
  vector<vector<ll>> A(n, vector<ll>(n, 0));
  for (int i = 0; i < n; ++i) A[i][i] = 1;
  return A;
}
vector<vector<ll>> matmult(vector<vector<ll>> A, vector<vector<ll>> B, ll mod) {
  int n = A.size();
  vector<vector<ll>> C(n, vector<ll>(n, 0));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      for (int k = 0; k < n; ++k) {
        C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod;
      }
    }
  }
  return C;
}
vector<vector<ll>> matpow(vector<vector<ll>> A, int e, ll mod) {
  if (e == 0) {
    return id((int)A.size());
  }
  auto H = matpow(A, e / 2, mod);
  H = matmult(H, H, mod);
  if (e % 2 == 0) {
    return H;
  }
  return matmult(H, A, mod);
}
long long egcd(long long a, long long b, long long& x, long long& y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  auto d = egcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long linear_modulo_diophantine(long long a, long long b, long long m) {
  if (0)
    cout << "Linear modulo solving " << a << " x = " << b << "  (mod " << m
         << ")\n";
  assert(a != 0);
  long long x, y;
  long long g = egcd(a, m, x, y);
  if (b % g != 0) return -1;
  x *= b / g;
  x %= m;
  if (x < 0) x += m;
  if (0) cout << "Returning  " << x << "\n";
  return x;
}
long long powmod(long long a, long long e, long long mod) {
  a %= mod;
  if (a < 0) a += mod;
  long long res = 1;
  while (e) {
    if (e & 1LL) res = res * a % mod;
    a = a * a % mod;
    e /= 2;
  }
  return res;
}
long long discrete_logarithm(long long h, long long g, const long long mod) {
  h %= mod;
  const auto m = static_cast<int>(ceil(sqrt(mod)));
  std::unordered_map<int, int> t;
  long long e = 1;
  for (int i = 0; i < m; ++i) {
    t[static_cast<int>(e)] = i;
    e = e * g % mod;
  }
  const auto factor = powmod(g, mod - m - 1, mod);
  e = h;
  for (int i = 0; i < m; ++i) {
    auto it = t.find(static_cast<int>(e));
    if (it != t.end()) return (1LL * m * i + 1LL * it->second) % mod;
    e = e * factor % mod;
  }
  return -1LL;
}
long long discrete_root(long long a, long long b, const long long g,
                        const long long p) {
  a %= p;
  b %= p;
  if (a == 0) return b == 1 ? -2 : -1;
  if (0) cout << "Discrete logarithm of " << b << " is ";
  b = discrete_logarithm(b, g, p);
  if (0) cout << b << "\n";
  long long logx = linear_modulo_diophantine(a, b, p - 1);
  if (logx == -1) return -1;
  return powmod(g, logx, p);
}
void solve() {
  const ll MOD = 998244353;
  int k;
  cin >> k;
  vector<ll> b(k);
  cin >> b;
  ll n, m;
  cin >> n >> m;
  --n;
  vector<vector<ll>> c(2 * k, vector<ll>(k, 0));
  for (int i = 0; i < k; ++i) c[i][i] = 1;
  for (int i = k; i < 2 * k; ++i)
    for (int j = 0; j < k; ++j)
      for (int l = 0; l < k; ++l)
        c[i][l] = (c[i][l] + c[i - j - 1][l] * b[j]) % (MOD - 1);
  vector<vector<ll>> A(k);
  for (int i = 0; i < k; ++i) A[i] = c[k + i];
  if (0) cout << "A: " << A << "\nn = " << n << "\n";
  A = matpow(A, n / k, MOD - 1);
  if (0) cout << "A: " << A << "\n";
  ll a = A[n % k][k - 1];
  if (0) cout << "a = " << a << "\n";
  ll ans = discrete_root(a, m, 3, MOD);
  if (ans == -2) ans = 1;
  cout << ans << "\n";
}