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;
inline long long read() {
  long long a = 0;
  char c = getchar();
  bool f = 0;
  while (!isdigit(c)) {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (isdigit(c)) {
    a = (a << 3) + (a << 1) + (c ^ '0');
    c = getchar();
  }
  return f ? -a : a;
}
const long long MOD = 998244353, G = 3;
long long K;
struct matrix {
  long long a[100][100];
  long long* operator[](long long x) { return a[x]; }
  matrix() { memset(a, 0, sizeof(a)); }
  matrix operator*(matrix b) {
    matrix c;
    for (long long i = 0; i < K; ++i)
      for (long long j = 0; j < K; ++j)
        for (long long k = 0; k < K; ++k)
          c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % (MOD - 1);
    return c;
  }
} S, T;
inline long long gcd(long long a, long long b) {
  long long r = a % b;
  while (r) {
    a = b;
    b = r;
    r = a % b;
  }
  return b;
}
inline long long poww(long long a, long long b, long long mod = MOD) {
  long long times = 1;
  while (b) {
    if (b & 1) times = times * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return times;
}
map<long long, long long> Hash;
inline long long BSGS(long long x) {
  long long t = sqrt(MOD) + 1, times = x;
  for (long long i = 0; i < t; i++) {
    Hash[times] = i;
    times = times * G % MOD;
  }
  times = poww(G, t);
  long long now = times;
  for (long long i = 1; i <= t + 1; i++) {
    if (Hash.count(now)) {
      return i * t - Hash[now];
    }
    now = now * times % MOD;
  }
  return -1;
}
long long phi(long long x) {
  long long times = x;
  for (long long i = 2; i * i <= x; ++i) {
    if (x % i == 0) {
      times = times / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  }
  if (x - 1) times = times / x * (x - 1);
  return times;
}
signed main() {
  K = read();
  for (long long i = 0; i < K; ++i) T[K - i - 1][K - 1] = read() % (MOD - 1);
  long long N = read() - K;
  long long t = BSGS(read());
  for (long long i = 0; i + 1 < K; ++i) T[i + 1][i] = 1;
  S[0][K - 1] = 1;
  while (N) {
    if (N & 1) S = S * T;
    T = T * T;
    N >>= 1;
  }
  long long cur = S[0][K - 1], p = gcd(cur, MOD - 1);
  if (t % p != 0)
    puts("-1");
  else {
    t /= p;
    cur /= p;
    long long mod = (MOD - 1) / p;
    cout << poww(G, poww(cur, phi(mod) - 1, mod) * t % mod);
  }
  return 0;
}