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;
using ll = long long;
const ll mod = 998244353;
ll sq = 31596;
struct Matrix {
  int n, m;
  vector<vector<ll>> a;
  Matrix(int n_, int m_) : n(n_), m(m_), a(n_, vector<ll>(m_)) {}
  vector<ll>& operator[](int i) { return a[i]; }
};
Matrix operator*(Matrix a, Matrix b) {
  Matrix res(a.n, b.m);
  for (int i = 0; i < res.n; i++)
    for (int j = 0; j < res.m; j++)
      for (int k = 0; k < a.m; k++)
        res[i][j] = (res[i][j] + a[i][k] * b[k][j]) % (mod - 1);
  return res;
}
Matrix BinPow(Matrix a, ll n) {
  if (n == 1) return a;
  if (n & 1) return a * BinPow(a, n - 1);
  return BinPow(a * a, n / 2);
}
ll Binpow(ll a, ll n) {
  if (n == 0) return 1;
  if (n & 1) return (a * Binpow(a, n - 1)) % mod;
  return Binpow((a * a) % mod, n / 2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int k;
  cin >> k;
  Matrix g(1, k), a(k, k);
  g[0][k - 1] = 1;
  for (int i = 0; i + 1 < k; i++) a[i + 1][i] = 1;
  for (int i = k - 1; i >= 0; i--) cin >> a[i][k - 1];
  ll n, m;
  cin >> n >> m;
  if (n - 1) {
    a = BinPow(a, n - 1);
    g = g * a;
  }
  ll t = Binpow(3, g[0][0]);
  vector<pair<ll, int>> q, p;
  for (ll i = 0; i < sq; i++) {
    ll tmp = Binpow(t, i);
    tmp = Binpow(tmp, mod - 2);
    tmp = (tmp * m) % mod;
    q.push_back({tmp, i});
  }
  sort(q.begin(), q.end());
  for (ll i = 0; i < mod / sq; i++)
    p.push_back({Binpow(t, (sq * i) % (mod - 1)), i});
  sort(p.begin(), p.end());
  int i = 0, j = 0;
  while (i < p.size() && j < q.size()) {
    while (i < p.size() && j < q.size() && p[i].first < q[j].first) i++;
    while (i < p.size() && j < q.size() && p[i].first > q[j].first) j++;
    if (i < p.size() && j < q.size() && p[i].first == q[j].first) {
      ll x = sq * p[i].second + q[j].second;
      cout << Binpow(3, x % (mod - 1));
      return 0;
    }
  }
  cout << -1;
  return 0;
}