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 int N = 103, M = 998244353;
struct matrix {
  long long a[N][N];
  matrix() { memset(a, 0, sizeof(a)); }
};
matrix operator*(matrix a, matrix b) {
  matrix ans;
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++)
      for (int k = 0; k < N; k++)
        ans.a[i][j] = (ans.a[i][j] + a.a[i][k] * b.a[k][j]) % (M - 1);
  return ans;
}
matrix power(matrix base, int k) {
  if (k == 1) return base;
  matrix ans = power(base, k / 2);
  ans = ans * ans;
  if (k & 1) ans = ans * base;
  return ans;
}
int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
long long powMod(long long a, long long b, long long p) {
  long long res = 1;
  while (b) {
    if (b & 1)
      res = (long long)(res * 1LL * a % p), --b;
    else
      a = (long long)(a * 1LL * a % p), b >>= 1;
  }
  return res;
}
long long generator(long long p) {
  vector<long long> fact;
  long long phi = p - 1, n = phi;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) {
      fact.push_back(i);
      while (n % i == 0) n /= i;
    }
  if (n > 1) fact.push_back(n);
  for (long long res = 2; res <= p; res++) {
    bool ok = true;
    for (size_t i = 0; i < fact.size() && ok; i++)
      ok &= powMod(res, phi / fact[i], p) != 1;
    if (ok) return res;
  }
  return -1;
}
long long discrete_log(long long a, long long b, long long m) {
  long long n = (long long)sqrt(m + .0) + 1;
  map<long long, long long> Hash;
  for (long long i = n; i >= 1; i--) Hash[powMod(a, i * n, m)] = i;
  for (long long i = 0; i <= n; i++) {
    long long cur = (powMod(a, i, m) * b) % m;
    if (Hash.count(cur)) {
      long long ans = Hash[cur] * n - i;
      if (ans < m) return ans;
    }
  }
  return -1;
}
long long discrete_root(long long a, long long b, long long p) {
  long long g = generator(p);
  assert(g != -1);
  long long k = discrete_log(powMod(g, a, p), b, p);
  if (k == -1) return -1;
  return powMod(g, k, p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  matrix ank;
  for (int i = 0; i < n; i++) cin >> ank.a[0][i];
  for (int j = 1; j < n; j++) ank.a[j][j - 1] = 1;
  int k, x;
  cin >> k >> x;
  matrix ans = power(ank, k - n);
  long long pw = ans.a[0][0];
  cout << discrete_root(pw, x, M) << endl;
  return 0;
}