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 p = 998244353;
struct Mat {
  int n, m;
  vector<long long> data;
  Mat(int _n, int _m) : n(_n), m(_m) { data.resize(n * m); }
  long long& operator()(int i, int j) { return data[i * m + j]; }
  const long long& operator()(int i, int j) const { return data[i * m + j]; }
  Mat operator*(const Mat& rhs) const {
    assert(m == rhs.n);
    Mat result(n, rhs.m);
    const Mat& lhs = *this;
    for (int i = 0; i < result.n; i++) {
      for (int j = 0; j < result.m; j++) {
        for (int k = 0; k < m; k++) {
          result(i, j) += lhs(i, k) * rhs(k, j);
          result(i, j) %= p - 1;
        }
      }
    }
    return result;
  }
  static Mat id(int n) {
    Mat result = Mat(n, n);
    for (int i = 0; i < n; i++) result(i, i) = 1;
    return result;
  }
  void print() const {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        fprintf(stderr, "%3lld ", this->operator()(i, j));
      }
      fprintf(stderr, "\n");
    }
  }
};
Mat mod_pow(Mat a, long long b) {
  Mat result = Mat::id(a.n);
  while (b > 0) {
    if (b % 2 == 1) {
      result = result * a;
    }
    a = a * a;
    b /= 2;
  }
  return result;
}
long long mod_pow(long long a, long long b) {
  long long result = 1;
  a %= p;
  while (b > 0) {
    if (b % 2 == 1) {
      result = result * a % p;
    }
    a = a * a % p;
    b /= 2;
  }
  return result;
}
long long log(long long a, long long b) {
  long long n = int(sqrt(p)) + 1;
  map<long long, long long> vals;
  for (int i = 1; i <= n; i++) vals[mod_pow(a, i * n)] = i;
  for (int i = 0; i <= n; i++) {
    long long cur = (mod_pow(a, i) * b) % p;
    if (vals.count(cur)) {
      long long ans = vals[cur] * n - i;
      return ans % (p - 1);
    }
  }
  return -1;
}
int main() {
  int k;
  scanf("%d", &k);
  Mat mat(k, k);
  for (int i = 0; i < k; i++) {
    scanf("%lld", &mat(0, i));
  }
  for (int i = 0; i < k - 1; i++) {
    mat(i + 1, i) = 1;
  }
  long long n, m;
  scanf("%lld %lld", &n, &m);
  Mat m2 = mod_pow(mat, n - k);
  Mat initial(k, 1);
  initial(0, 0) = 1;
  long long pow_of_fk = (m2 * initial)(0, 0);
  fprintf(stderr, "pow_of_fk=%lld\n", pow_of_fk);
  long long base = mod_pow(3, pow_of_fk);
  long long dlog = log(base, m);
  fprintf(stderr, "base = %lld, m = %lld, dlog = %lld\n", base, m, dlog);
  if (dlog == -1) {
    puts("-1");
    return 0;
  }
  printf("%lld\n", mod_pow(3, dlog));
  return 0;
}