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;
vector<vector<long long>> mul(const vector<vector<long long>> &a,
                              const vector<vector<long long>> &b,
                              long long mod) {
  int n = a.size(), m = b[0].size();
  vector<vector<long long>> res(n, vector<long long>(m, 0));
  int mid = a[0].size();
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      for (int k = 0; k < mid; k++) {
        (res[i][j] += a[i][k] * b[k][j]) %= mod;
      }
    }
  }
  return res;
}
vector<vector<long long>> unit(int n) {
  vector<vector<long long>> res(n, vector<long long>(n, 0));
  for (int i = 0; i < n; i++) res[i][i] = 1;
  return res;
}
vector<vector<long long>> pow(const vector<vector<long long>> &a, long long b,
                              long long mod) {
  vector<vector<long long>> res = unit(a.size()), c = a;
  while (b > 0) {
    if (b & 1) res = mul(res, c, mod);
    c = mul(c, c, mod);
    b /= 2;
  }
  return res;
}
long long mod_pow(long long a, long long b, long long mod) {
  long long res = 1, c = a;
  while (b) {
    if (b & 1) res = res * c % mod;
    c = c * c % mod;
    b /= 2;
  }
  return res;
}
long long discrete_logarithm(long long a, long long b, long long mod) {
  long long sq = sqrt((long double)mod);
  vector<pair<long long, int>> arr;
  long long c = a;
  for (int i = 0; i < sq; i++) {
    long long v = b * c % mod;
    c = c * a % mod;
    arr.push_back(make_pair(v, sq - i - 1));
  }
  sort(arr.begin(), arr.end());
  long long as = mod_pow(a, sq, mod);
  c = as;
  for (int i = 0; i <= (mod / sq); i++) {
    long long v = c;
    int ind =
        lower_bound(arr.begin(), arr.end(), make_pair(v, -1)) - arr.begin();
    while (ind < arr.size() && v == arr[ind].first) {
      long long res = i * sq + arr[ind].second;
      if (b == mod_pow(a, res, mod)) return res;
      ind++;
    }
    c = c * as % mod;
  }
  return -1;
}
long long inv(long long a, long long mod) {
  long long b = mod, p = 1, q = 0;
  while (b) {
    long long c = a / b, d = a;
    a = b;
    b = d % b;
    d = p;
    p = q;
    q = d - c * q;
  }
  return (p + mod) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int k;
  cin >> k;
  vector<long long> b(k);
  for (int i = 0; i < k; i++) cin >> b[i];
  long long mod = 998244353;
  vector<vector<long long>> mat(k, vector<long long>(k));
  for (int i = 0; i < k - 1; i++) {
    mat[i][i + 1] = 1;
  }
  for (int i = 0; i < k; i++) {
    mat[k - 1][i] = b[k - 1 - i];
  }
  int n;
  long long v;
  cin >> n >> v;
  n--;
  mat = pow(mat, n, mod - 1);
  if (mat[0][k - 1] == 0) {
    if (v == 1)
      cout << 1 << "\n";
    else
      cout << -1 << "\n";
    return 0;
  }
  long long ex = discrete_logarithm(3, v, mod);
  long long im = inv(mat[0][k - 1], mod - 1);
  long long g = mat[0][k - 1] * im % (mod - 1);
  long long z = im * (ex / g) % (mod - 1);
  long long x = mod_pow(3, z, mod);
  if (v == mod_pow(x, mat[0][k - 1], mod))
    cout << x << "\n";
  else
    cout << -1 << "\n";
  return 0;
}