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 maxN = 1e2 + 9, maxV = 200 + 9, MOD = 998244353, AND = (1 << 20) - 1,
          lg = 20;
int n, k, m;
vector<vector<long long>> operator*(const vector<vector<long long>> &one,
                                    const vector<vector<long long>> &two) {
  vector<vector<long long>> ret(one.size(), vector<long long>(two[0].size()));
  for (int i = 0; i < one.size(); i++)
    for (int j = 0; j < two[0].size(); j++)
      for (int u = 0; u < two.size(); u++)
        ret[i][j] = (ret[i][j] + one[i][u] * two[u][j]) % (MOD - 1);
  return ret;
}
vector<vector<long long>> operator^(vector<vector<long long>> a, int b) {
  vector<vector<long long>> ret(k, vector<long long>(k));
  for (int i = 0; i < k; i++) ret[i][i] = 1;
  while (b) {
    if (b & 1) ret = ret * a;
    b >>= 1;
    a = a * a;
  }
  return ret;
}
long long fastPow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ret;
}
int solv(long long a, long long b) {
  const int limit = 40000;
  map<int, int> mp;
  for (int i = 0; i <= limit; i++) mp[b * fastPow(a, i) % MOD] = i;
  for (int i = 1; i < MOD + limit; i += limit)
    if (mp.find((int)fastPow(a, i)) != mp.end())
      return (MOD - 1 + i - mp[(int)fastPow(a, i)]) % (MOD - 1);
  return -1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> k;
  vector<vector<long long>> base(k, vector<long long>(k));
  for (int i = 0; i < k; i++) cin >> base[0][i], base[0][i] %= MOD - 1;
  for (int i = 1; i < k; i++) base[i][i - 1] = 1;
  cin >> n >> m;
  n -= k;
  vector<vector<long long>> start(k, vector<long long>({0}));
  start[0][0] = 1;
  vector<vector<long long>> ending = (base ^ n) * start;
  long long ende = ending[0][0];
  int logar = solv(fastPow(3, ende), m);
  if (logar == -1) {
    cout << "-1\n";
    return 0;
  }
  cout << fastPow(3, logar) << '\n';
}