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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
const int INF = 1 << 29;
const int MOD = 998244353;
long long gcd(long long a, long long b) {
  while (b) {
    long long c = a;
    a = b;
    b = c % b;
  }
  return a;
}
long long extended_gcd(long long a, long long b, long long& x, long long& y) {
  long long t;
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long d = extended_gcd(b, a % b, x, y);
  t = x, x = y;
  y = t - a / b * y;
  return d;
}
long long power_mod(long long a, int n, long long p) {
  long long ret = 1;
  for (; n; n >>= 1) {
    if (n & 1) ret = ret * a % p;
    a = a * a % p;
  }
  return ret;
}
int baby(long long a, long long p, long long b) {
  int m = 1 + (int)(sqrt((double)p));
  long long x, y, prod = 1;
  unordered_map<long long, int> A;
  for (int i = 0; i < m; ++i) {
    A[prod] = i;
    prod = prod * a % p;
  }
  extended_gcd(a, p, x, y);
  x %= p;
  if (x < 0) x += p;
  x = power_mod(x, m, p);
  prod = b;
  for (int i = 0; i < m; ++i) {
    auto it = A.find(prod);
    if (it != A.end()) return i * m + it->second;
    prod = prod * x % p;
  }
  return -INF;
}
int discrete_log(long long a, long long p, long long b) {
  if (b == 1) return 0;
  long long d = gcd(a, p);
  if (d == 1) return baby(a, p, b);
  if (b % d) return -1;
  b /= d;
  p /= d;
  long long x, y;
  extended_gcd(a / d, p, x, y);
  x %= p;
  if (x < 0) x += p;
  b = b * x % p;
  return 1 + discrete_log(a, p, b);
}
const int N = 100;
int a[N][N], b[N], t[N][N], ret[N][N];
void mul(int a[N][N], int b[N][N], int c[N][N], int n, int mod) {
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      c[i][j] = 0;
      for (int k = 0; k < n; ++k) {
        c[i][j] += (long long)a[i][k] * b[k][j] % mod;
        c[i][j] %= mod;
      }
    }
  }
}
int main() {
  int k, n, m;
  scanf("%d", &k);
  for (int i = 0; i < k; ++i) {
    scanf("%d", &b[i]);
  }
  for (int i = 0; i < k - 1; ++i) a[i][i + 1] = 1;
  for (int i = 0; i < k; ++i) a[k - 1][i] = b[k - 1 - i];
  scanf("%d%d", &n, &m);
  n -= k;
  m = discrete_log(3, MOD, m);
  for (int i = 0; i < k; ++i) ret[i][i] = 1;
  for (; n; n >>= 1) {
    if (n & 1) {
      mul(ret, a, t, k, MOD - 1);
      memcpy(ret, t, sizeof(ret));
    }
    mul(a, a, t, k, MOD - 1);
    memcpy(a, t, sizeof(a));
  }
  long long A = ret[k - 1][k - 1], B = m;
  long long x, y;
  long long d = extended_gcd(A, MOD - 1, x, y);
  if (B % d) {
    puts("-1");
    return 0;
  }
  x %= MOD - 1;
  if (x < 0) x += MOD - 1;
  x = B / d * x % (MOD - 1);
  printf("%lld\n", power_mod(3, x, MOD));
  return 0;
}