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>
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline void sub(long long &a, long long b) {
  a -= b;
  if (a < 0) a += 998244353;
}
inline void add(long long &a, long long b) {
  a += b;
  if (a >= 998244353) a -= 998244353;
}
inline long long mul(long long a, long long b, long long c) {
  return (a * b - (long long)((long double)a * b / c) * c + c) % c;
}
inline long long qp(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 998244353;
    a = a * a % 998244353, b >>= 1;
  }
  return ans;
}
inline long long qp(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a, c);
    a = mul(a, a, c), b >>= 1;
  }
  return ans;
}
using namespace std;
const long double pi = acos(-1);
const unsigned long long ba = 233;
const double eps = 1e-8;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int N = 100000 + 10, maxn = 1000000 + 10, inf = 0x3f3f3f3f;
int k;
struct info {
  long long a[110][110];
  void cl() { memset(a, 0, sizeof a); }
  void init() {
    cl();
    for (int i = 0; i < k; i++) a[i][i] = 1;
  }
  void pr() {
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) printf("%d ", a[i][j]);
      puts("");
    }
  }
};
long long b[N];
info mul(info a, info b) {
  info c;
  c.cl();
  for (int i = 0; i < k; i++)
    for (int j = 0; j < k; j++)
      for (int l = 0; l < k; l++)
        c.a[i][l] = (c.a[i][l] + a.a[i][j] * b.a[j][l]) % (998244353 - 1);
  return c;
}
info fp(info a, long long b) {
  info c;
  c.init();
  while (b) {
    if (b & 1) c = mul(c, a);
    a = mul(a, a);
    b >>= 1;
  }
  return c;
}
long long bsgs(long long a, long long b, long long p) {
  if (a % p == 0) return -1;
  map<long long, long long> ma;
  long long m = ceil(sqrt(p)), now = b % p;
  ma[now] = 0;
  for (int i = 1; i <= m; i++) {
    now = now * a % p;
    ma[now] = i;
  }
  long long ans = -1, t = qp(a, m, p);
  now = 1;
  for (int i = 1; i <= m; i++) {
    now = now * t % p;
    if (ma[now]) {
      ans = i * m - ma[now];
      break;
    }
  }
  if (ans != -1)
    return (ans % p + p) % p;
  else
    return -1;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  long long ans = exgcd(b, a % b, x, y);
  long long t = x;
  x = y;
  y = t - a / b * y;
  return ans;
}
int main() {
  scanf("%d", &k);
  for (int i = 0; i < k; i++) scanf("%lld", &b[i]);
  long long n, m;
  scanf("%lld%lld", &n, &m);
  info A;
  A.cl();
  for (int i = 0; i < k; i++) A.a[0][i] = b[i];
  for (int i = 1; i < k; i++) A.a[i][i - 1] = 1;
  A = fp(A, n - k);
  long long a = A.a[0][0], b = 998244353 - 1, c = bsgs(3, m, 998244353), x, y;
  long long r = gcd(a, b);
  if (c % r != 0) return 0 * puts("-1");
  exgcd(a, b, x, y);
  x *= c / r;
  long long p = b / r;
  x = (x % p + p) % p;
  printf("%lld\n", qp(3, x));
  return 0;
}