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;
struct Mat {
  long long m[110][110];
  Mat() {
    for (int i = 1; i <= 101; i++)
      for (int j = 1; j <= 101; j++) m[i][j] = 0;
  }
};
Mat a, e;
long long N;
Mat Mul(Mat &x, Mat &y) {
  Mat c;
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= N; j++)
      for (int k = 1; k <= N; k++) {
        c.m[i][j] = (c.m[i][j] + x.m[i][k] * y.m[k][j] % (998244353 - 1)) %
                    (998244353 - 1);
      }
  return c;
}
Mat pow(Mat x, long long y) {
  Mat ans = e;
  while (y) {
    if (y & 1) ans = Mul(ans, x);
    x = Mul(x, x);
    y >>= 1;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline long long fast_exp(long long x, long long n, long long p) {
  long long z = 1;
  for (long long y = x % p; n; y = y * y % p, n >>= 1)
    if (n & 1) z = z * y % p;
  return z;
}
long long exbsgs(long long a, long long y, long long p) {
  a %= p;
  y %= p;
  if (a == 0) return y > 1 ? -1 : y == 0 && p > 1;
  long long g, c = 0, q = 1;
  while ((g = gcd(a, p)) != 1) {
    if (y == 1) return c;
    if (y % g) return -1;
    ++c;
    y /= g;
    p /= g;
    q = a / g * q % p;
  }
  map<long long, long long> x;
  long long m = sqrt(p);
  for (long long i = 1, t = y * a % p; i <= m; ++i, t = t * a % p) x[t] = i;
  for (long long i = m, t = fast_exp(a, m, p); i - m < p - 1; i += m)
    if (q = q * t % p, x.count(q)) return i - x[q] + c;
  return -1;
}
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> a.m[1][i + 1];
    a.m[i + 2][i + 1] = 1;
    e.m[i + 1][i + 1] = 1;
  }
  long long k, m;
  cin >> k >> m;
  Mat ac = pow(a, k - N);
  long long top = ac.m[1][1];
  long long g = fast_exp(3, top, 998244353);
  long long ans = exbsgs(g, m, 998244353);
  if (ans == -1) {
    cout << "-1" << endl;
    return 0;
  }
  ans = fast_exp(3, ans, 998244353);
  cout << ans << endl;
  return 0;
}