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, p1 = p - 1;
long long k, n, fn, f[111], b[111];
long long mod_pow(long long a, long long k, long long MOD);
vector<long long> factor(long long n) {
  vector<long long> res;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      res.push_back(i);
      if (i * i != n) {
        res.push_back(n / i);
      }
    }
  }
  return res;
}
bool is_primitive_root(long long g, long long n) {
  auto fc = factor(n - 1);
  for (auto x : fc) {
    if (mod_pow(g, x, n) <= 1) return false;
  }
  return true;
}
long long primitive_root(long long n) {
  while (1) {
    long long g = rand() % (n - 1) + 1;
    if (is_primitive_root(g, n)) return g;
  }
  return -1;
}
long long g;
void gcd(long long a, long long b, long long &d, long long &x, long long &y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    gcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long inv(long long a, long long n) {
  long long d, x, y;
  gcd(a, n, d, x, y);
  return d == 1 ? (x + n) % n : -1;
}
long long mod_pow(long long a, long long k, long long MOD) {
  long long res = 1;
  while (k) {
    if (k & 1) res = res * a % MOD;
    a = a * a % MOD;
    k /= 2;
  }
  return res;
}
long long mod_log(long long a, long long b, long long n) {
  long long m = sqrt(n + 0.5), v = inv(mod_pow(a, m, n), n), e = 1;
  map<long long, long long> x;
  x[1] = 0;
  for (long long i = 1; i < m; i++) {
    e = e * a % n;
    if (!x.count(e)) x[e] = i;
  }
  for (long long i = 0; i < n; i += m) {
    if (x.count(b)) return i + x[b];
    b = b * v % n;
  }
  return -1;
}
vector<vector<long long> > mul(vector<vector<long long> > &A,
                               vector<vector<long long> > &B) {
  vector<vector<long long> > C(::k, vector<long long>(::k));
  for (int i = 0; i < ::k; i++) {
    for (int k = 0; k < ::k; k++) {
      for (int j = 0; j < ::k; j++) {
        C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % p1;
      }
    }
  }
  return C;
}
vector<vector<long long> > pow(vector<vector<long long> > A, long long n) {
  vector<vector<long long> > B(k, vector<long long>(k));
  for (int i = 0; i < k; i++) B[i][i] = 1;
  while (n) {
    if (n & 1) B = mul(B, A);
    A = mul(A, A);
    n >>= 1;
  }
  return B;
}
long long mod_fun(long long a, long long b, long long n) {
  long long x, y, d;
  gcd(a, n, d, x, y);
  if (b % d) return -1;
  x = x * (b / d) % n;
  x = (x + n) % n;
  return (x % (n / d) + (n / d)) % (n / d);
}
int main() {
  srand(time(NULL));
  ios::sync_with_stdio(0);
  cin >> k;
  for (int i = 0; i < k; i++) cin >> b[i];
  cin >> n >> fn;
  g = primitive_root(p);
  long long hn = mod_log(g, fn, p);
  vector<vector<long long> > A(k, vector<long long>(k));
  for (int i = 0; i < k; i++) A[0][i] = b[i] % p1;
  for (int i = 1; i < k; i++) A[i][i - 1] = 1;
  A = pow(A, n - k);
  long long c = A[0][0];
  long long hk = mod_fun(c, hn, p1);
  if (hk < 0) {
    cout << -1 << endl;
    return 0;
  }
  long long fk = mod_pow(g, hk, p);
  cout << fk << endl;
  return 0;
}