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 int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 998244353;
const int maxn = 105;
int N, M, K, b[maxn];
struct Ma {
  long long m[maxn][maxn];
  Ma() { clear(); }
  void clear() { memset((m), 0, sizeof(m)); }
  void init() {
    for (int i = 1; i <= K; i++) m[i][i] = 1;
  }
  Ma operator*(Ma &T) {
    Ma re;
    for (int i = 1; i <= K; i++)
      for (int j = 1; j <= K; j++)
        for (int k = 1; k <= K; k++)
          re.m[i][j] += m[i][k] * T.m[k][j] % (mod - 1),
              re.m[i][j] %= (mod - 1);
    return re;
  }
};
long long qp(long long x, long long n) {
  long long re = 1;
  while (n) {
    if (n & 1) re = re * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return re;
}
long long BSGS(long long a, long long b) {
  map<long long, long long> tab;
  long long u = (long long)ceil(sqrt(mod));
  long long now = 1, step;
  for (long long i = 0; i < u; i++) {
    long long tem = b * qp(now, mod - 2) % mod;
    if (!tab.count(tem)) tab[tem] = i;
    now = now * a % mod;
  }
  step = now;
  now = 1;
  for (int i = 0; i < mod; i += u) {
    if (tab.count(now)) return i + tab[now];
    now = now * step % mod;
  }
  return -1;
}
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long re = gcd(b, a % b, y, x);
  y -= a / b * x;
  return re;
}
long long solve2(long long p) {
  long long b = BSGS(3, M);
  long long x, y, d = gcd(p, mod - 1, x, y);
  if (b % d) return -1;
  x = (x * b / d % (mod - 1) + mod - 1) % (mod - 1);
  return qp(3, x);
}
Ma qp(Ma t, long long n) {
  Ma re;
  re.init();
  while (n) {
    if (n & 1) re = re * t;
    t = t * t;
    n >>= 1;
  }
  return re;
}
long long solve1() {
  Ma A;
  for (int i = 1; i <= K - 1; i++) A.m[i + 1][i] = 1;
  for (int i = 1; i <= K; i++) A.m[i][K] = b[K - i + 1];
  A = qp(A, N - 1);
  return A.m[K][1];
}
int main() {
  scanf("%d", &K);
  for (int i = 1; i <= K; i++) scanf("%d", &b[i]);
  scanf("%d%d", &N, &M);
  long long p = solve1();
  printf("%d\n", solve2(p));
}