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 <typename T>
T binpow(T a, T n, T m) {
  T res = 1;
  for (; n > 0; n /= 2) {
    if (n & 1) {
      res = (res * a) % m;
    }
    a = (a * a) % m;
  }
  return res;
}
template <typename T>
std::pair<bool, T> discr_log(T a, T b, T m) {
  T n = (T)std::sqrt(double(m));
  std::map<T, size_t> fn;
  T an0 = binpow(a, n, m);
  T an = an0;
  for (size_t i = 0; i < (m + n - 1) / n; ++i) {
    if (fn.count(an) > 0)
      break;
    else {
      fn[an] = i + 1;
      an = (an * an0) % m;
    }
  }
  T aq = 1;
  for (size_t i = 0; i < n; ++i, aq = (aq * a) % m) {
    if (fn.count((aq * b) % m) > 0) return {true, n * fn[(aq * b) % m] - i};
  }
  return {false, 0};
}
template <typename T>
T first_order_root_prime(T m) {
  T fi = m - 1;
  std::vector<T> p;
  T cur = fi;
  for (size_t i = 2; i * i <= fi; ++i) {
    if (cur % i == 0) {
      p.push_back(fi / i);
      for (; cur % i == 0; cur /= i)
        ;
    }
  }
  if (cur > 1) p.push_back(fi / cur);
  for (T i = 2; i < m; ++i) {
    bool g = true;
    for (size_t j = 0; j < p.size(); ++j)
      if (binpow(i, p[j], m) == 1) {
        g = false;
        break;
      }
    if (g) return i;
  }
  return 0;
}
template <typename T>
std::pair<bool, T> discr_sqrt_prime(T b, T a, T m) {
  T fo = first_order_root_prime(m);
  T c = binpow(fo, b, m);
  std::pair<bool, T> q = discr_log(c, a, m);
  return {q.first, binpow(fo, q.second, m)};
}
template <typename T>
struct matr {
  matr(std::vector<T> const& b, T m_)
      : n(b.size()),
        m(m_),
        a(std::vector<std::vector<T>>(n, std::vector<T>(n, 0))) {
    for (size_t i = 0; i < n; ++i) a[i][0] = b[i];
    for (size_t i = 0; i + 1 < n; ++i) a[i][i + 1] = 1;
  }
  matr(size_t n_, T m_)
      : n(n_), m(m_), a(std::vector<std::vector<T>>(n, std::vector<T>(n, 0))) {}
  matr operator*(matr const& other) {
    matr res(n, m);
    for (size_t i = 0; i < n; ++i)
      for (size_t k = 0; k < n; ++k)
        for (size_t j = 0; j < n; ++j)
          res.a[i][j] = (res.a[i][j] + a[i][k] * other.a[k][j]) % m;
    return res;
  }
  T get(size_t i, size_t j) { return a[i][j]; }
  void print() {
    std::cout << "n = " << n << "\n";
    for (size_t i = 0; i < n; ++i) {
      for (size_t j = 0; j < n; ++j) std::cout << a[i][j] << " ";
      std::cout << "\n";
    }
  }
  matr binpow(T b) {
    matr res(n, m);
    matr c = *this;
    for (size_t i = 0; i < n; ++i) res.a[i][i] = 1;
    for (; b > 0; b /= 2) {
      if (b & 1) {
        res = res * c;
      }
      c = c * c;
    }
    return res;
  }
  int n;
  T m;
  std::vector<std::vector<T>> a;
};
int main() {
  int k;
  const long long MOD = 998244353LL;
  std::cin >> k;
  std::vector<long long> b(k);
  for (int i = 0; i < k; ++i) std::cin >> b[i];
  matr<long long> a(b, MOD - 1);
  long long n, m;
  std::cin >> n >> m;
  matr<long long> c = a.binpow(n - k);
  std::pair<bool, long long> res = discr_sqrt_prime(c.get(0, 0), m, MOD);
  if (res.first)
    std::cout << res.second << "\n";
  else
    std::cout << "-1\n";
}