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<long long> factors;
int IT_MAX = 1 << 17;
int MOD = 998244353;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double ERR = 1e-10;
long long mod(long long a, long long m) {
  return (a + (abs(a) / m) * m + m) % m;
}
vector<vector<long long>> b;
vector<long long> f;
vector<vector<long long>> mul(vector<vector<long long>>& a,
                              vector<vector<long long>>& b) {
  vector<vector<long long>> res(a.size(), vector<long long>(a.size(), 0));
  for (int i = 0; i < a.size(); ++i) {
    for (int j = 0; j < a.size(); ++j) {
      for (int k = 0; k < a.size(); ++k) {
        res[i][j] += a[i][k] * b[k][j];
        res[i][j] %= (MOD - 1);
      }
    }
  }
  return res;
}
vector<vector<long long>> binpow(vector<vector<long long>>& a, int n) {
  vector<vector<long long>> res(a.size(), vector<long long>(a.size(), 0));
  for (int i = 0; i < a.size(); ++i) {
    res[i][i] = 1;
  }
  while (n) {
    if (n & 1) res = mul(res, a);
    a = mul(a, a);
    n >>= 1;
  }
  return res;
}
long long bp(long long a, int n) {
  long long res = 1;
  while (n) {
    if (n & 1) {
      res *= a;
      res %= MOD;
    }
    a *= a;
    a %= MOD;
    n >>= 1;
  }
  return res;
}
long long solve(long long a, long long b, long long m) {
  long long n = (long long)sqrt(m + .0) + 1;
  long long an = 1;
  for (long long i = 0; i < n; ++i) an = (an * a) % m;
  map<long long, long long> vals;
  for (long long i = 1, cur = an; i <= n; ++i) {
    if (!vals.count(cur)) vals[cur] = i;
    cur = (cur * an) % m;
  }
  for (long long i = 0, cur = b; i <= n; ++i) {
    if (vals.count(cur)) {
      long long ans = vals[cur] * n - i;
      if (ans < m) return ans;
    }
    cur = (cur * a) % m;
  }
  return -1;
}
int generator(long long MOD) {
  vector<long long> fact;
  long long phi = MOD - 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 <= MOD; ++res) {
    bool ok = true;
    for (size_t i = 0; i < fact.size() && ok; ++i)
      ok &= bp(res, phi / fact[i]) != 1;
    if (ok) return res;
  }
  return -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  long long k;
  cin >> k;
  b.assign(k, vector<long long>(k, 0));
  for (int i = 0; i < k; ++i) {
    cin >> b[0][i];
  }
  for (int i = 0; i < k - 1; ++i) {
    b[i + 1][i] = 1;
  }
  long long n, m;
  cin >> n >> m;
  b = binpow(b, n - k);
  long long deg = b[0][0];
  long long per = generator(MOD);
  long long osn = bp(3, deg);
  if (osn == 0 || osn % MOD == 0) {
    cout << -1;
    return 0;
  }
  long long ndeg = solve(osn, m, MOD);
  if (ndeg == -1) {
    cout << -1;
    return 0;
  }
  cout << bp(3, ndeg);
  return 0;
}