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;
long long quick_exp(long long a, int n, int mod) {
  if (n == 0) return 1;
  long long q = quick_exp(a, n / 2, mod);
  q = (q * q) % mod;
  return (n % 2 ? (a * q) % mod : q);
}
int discrete_log(int g, int c, int mod) {
  int m = sqrt(mod);
  if (m * m < mod) m++;
  unordered_map<int, int> table;
  long long g_pow = 1;
  for (int j = 0; j < m; j++) {
    table[g_pow] = j;
    g_pow = (g_pow * g) % mod;
  }
  g_pow = quick_exp(g, mod - 1 - m, mod);
  int trial = c;
  for (int i = 0; i < m; i++) {
    unordered_map<int, int>::iterator it = table.find(trial);
    if (it != table.end()) return i * m + it->second;
    trial = (trial * g_pow) % mod;
  }
}
int cong(int a, int b, int mod) {
  int a1 = mod, b1 = 0;
  int a2 = a % mod, b2 = b % mod;
  while (0 < a2) {
    long long c = a1 / a2;
    a1 -= c * a2;
    b1 = ((long long)mod * mod + b1 - c * b2) % mod;
    swap(a1, a2);
    swap(b1, b2);
  }
  if (b2 != 0) return -1;
  return b1 / a1;
}
void matrix_mult(int A[100][100], int B[100][100], int k, int mod) {
  long long P[100][100], Q[100][100];
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < k; j++) {
      P[i][j] = A[i][j];
      Q[i][j] = B[i][j];
    }
  }
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < k; j++) {
      long long s = 0;
      for (int r = 0; r < k; r++) s = (s + P[i][r] * Q[r][j]) % mod;
      A[i][j] = s;
    }
  }
}
const int p = 998244353;
int k, A[100][100], X[100][100], n, m;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> k;
  for (int i = 0; i < k - 1; i++) A[i][i + 1] = 1;
  for (int i = 0; i < k; i++) cin >> A[k - 1][k - 1 - i];
  for (int i = 0; i < k; i++) X[i][i] = 1;
  cin >> n >> m;
  n--;
  while (n) {
    if (n % 2) matrix_mult(X, A, k, p - 1);
    matrix_mult(A, A, k, p - 1);
    n /= 2;
  }
  int a = X[0][k - 1];
  int b = discrete_log(3, m, p);
  int x = cong(a, b, p - 1);
  if (x == -1)
    cout << -1 << endl;
  else
    cout << quick_exp(3, x, p) << endl;
  return 0;
}