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;
const long long mod = p - 1;
long long k;
vector<long long> b;
long long n, m;
vector<vector<long long>> identity() {
  vector<vector<long long>> I(k, vector<long long>(k));
  for (int i = 0; i < k; i++) I[i][i] = 1;
  return I;
}
vector<vector<long long>> operator*(const vector<vector<long long>>& A,
                                    const vector<vector<long long>>& B) {
  vector<vector<long long>> C(k, vector<long long>(k));
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < k; j++) {
      for (int l = 0; l < k; l++) {
        C[i][j] += A[i][l] * B[l][j] % mod;
        C[i][j] %= mod;
      }
    }
  }
  return C;
}
vector<vector<long long>> pow(vector<vector<long long>> A, long long b) {
  vector<vector<long long>> result = identity();
  for (; b > 0; b /= 2) {
    if (b % 2) result = result * A;
    A = A * A;
  }
  return result;
}
long long modpow(long long a, long long b, long long mod) {
  long long result = 1;
  for (; b > 0; b /= 2) {
    if (b % 2) result = result * a % mod;
    a = a * a % mod;
  }
  return result;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
struct cp_algo {
  int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
  int powmod(int a, int b, int p) {
    int res = 1;
    while (b)
      if (b & 1)
        res = int(res * 1ll * a % p), --b;
      else
        a = int(a * 1ll * a % p), b >>= 1;
    return res;
  }
  int generator(int p) {
    vector<int> fact;
    int phi = p - 1, n = phi;
    for (int 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 (int 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;
  }
  int discrete_root(int n, int k, int a) {
    if (a == 0) return 0;
    int g = generator(n);
    int sq = (int)sqrt(n + .0) + 1;
    vector<pair<int, int>> dec(sq);
    for (int i = 1; i <= sq; ++i)
      dec[i - 1] = make_pair(powmod(g, int(i * sq * 1ll * k % (n - 1)), n), i);
    sort(dec.begin(), dec.end());
    int any_ans = -1;
    for (int i = 0; i < sq; ++i) {
      int my = int(powmod(g, int(i * 1ll * k % (n - 1)), n) * 1ll * a % n);
      vector<pair<int, int>>::iterator it =
          lower_bound(dec.begin(), dec.end(), make_pair(my, 0));
      if (it != dec.end() && it->first == my) {
        any_ans = it->second * sq - i;
        break;
      }
    }
    if (any_ans == -1) return -1;
    int delta = (n - 1) / gcd(k, n - 1);
    return powmod(g, any_ans % delta, n);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin >> k;
  vector<vector<long long>> A(k, vector<long long>(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];
  cin >> n >> m;
  A = pow(A, n - k);
  long long x = A[k - 1][k - 1];
  cp_algo c;
  cout << c.discrete_root(p, x, m) << endl;
}