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>
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 998244353;
const int root = 3;
int trans[100100];
std::map<int, int> first;
int solve(int en) {
  if (root == en) return 1;
  if (en == 1) return 0;
  int step = sqrt(MOD) + 10;
  trans[0] = 1;
  for (int i = 1; i <= step; i++) {
    trans[i] = (long long)trans[i - 1] * root % MOD;
    if (trans[i] == en) return i;
  }
  if (first.empty()) {
    int lst = 0;
    int val = 1;
    for (int i = 0; lst - step <= MOD; i++) {
      if (!first.count(val)) {
        first[val] = lst;
      }
      lst += step;
      val = (long long)val * trans[step] % MOD;
    }
  }
  for (int i = 0; i <= step; i++) {
    if (first.count(en)) {
      return (MOD - 1 - i + first[en]) % (MOD - 1);
    }
    en = (long long)en * root % MOD;
  }
  assert(false);
  return -1;
}
template <int mod = MOD>
struct modBase {
  modBase(int val = 0) : val(val) {}
  int val;
  modBase<mod> operator*(modBase<mod> o) {
    return (long long)val * o.val % mod;
  }
  modBase<mod> operator+(modBase<mod> o) {
    return val + o.val > mod ? val + o.val - mod : val + o.val;
  }
};
template <const size_t n, const size_t m, class T = modBase<>>
struct Matrix {
  T v[n][m];
  Matrix(int d = 0) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        v[i][j] = T(0);
      }
      if (i < m) {
        v[i][i] = T(d);
      }
    }
  }
  template <size_t mm>
  Matrix<n, mm, T> operator*(Matrix<m, mm, T> &o) {
    Matrix<n, mm, T> ans;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < mm; j++) {
        for (int k = 0; k < m; k++) {
          ans.v[i][j] = ans.v[i][j] + v[i][k] * o.v[k][j];
        }
      }
    }
    return ans;
  }
};
template <class T>
T fexp(T x, long long e) {
  T ans(1);
  for (; e > 0; e /= 2) {
    if (e & 1) ans = ans * x;
    x = x * x;
  }
  return ans;
}
long long modinverse(long long a, long long b, long long s0 = 1,
                     long long s1 = 0) {
  if (!b)
    return s0;
  else
    return modinverse(b, a % b, s1, s0 - s1 * (a / b));
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int k;
  std::cin >> k;
  std::vector<int> b(k);
  for (int i = 0; i < k; i++) {
    std::cin >> b[i];
  }
  Matrix<100, 100, modBase<MOD - 1>> mat;
  for (int i = 0; i < k; i++) {
    mat.v[0][i] = b[i];
  }
  for (int i = 1; i < k; i++) {
    mat.v[i][i - 1] = 1;
  }
  int n, val;
  std::cin >> n >> val;
  auto end = fexp(mat, n - k);
  int tot = end.v[0][0].val;
  int id = solve(val);
  long long g = gcd(id, tot);
  if (g == 0) g = 1;
  long long x = (id / g) * modinverse(tot / g, MOD - 1) % (MOD - 1);
  x = (x + MOD - 1) % (MOD - 1);
  long long ans = fexp(modBase<MOD>(root), x).val;
  if (fexp(modBase<MOD>(ans), tot).val == val) {
    std::cout << ans << '\n';
  } else {
    std::cout << "-1\n";
  }
}