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;
int N, M, K, a, n;
map<int, int> mp;
struct Matrix {
  long long a[100][100];
  inline long long* operator[](int x) { return a[x]; }
  Matrix operator*(Matrix t) {
    Matrix ans;
    memset(ans.a, 0, sizeof ans.a);
    for (int i = 0; i < K; ++i)
      for (int k = 0; k < K; ++k)
        for (int j = 0; j < K; ++j)
          ans[i][j] = (ans[i][j] + a[i][k] * t[k][j]) % 998244352;
    return ans;
  }
  Matrix Pow(int x) {
    Matrix cur(*this), ans;
    for (int i = 0; i < K; ++i) ans[i][i] = 1;
    for (; x; x >>= 1, cur = cur * cur)
      if (x & 1) ans = ans * cur;
    return ans;
  }
} t, ans;
long long Pow(long long x, long long y) {
  long long ans(1);
  for (; y; y >>= 1, x = x * x % 998244353)
    if (y & 1) ans = ans * x % 998244353;
  return ans;
}
int main() {
  scanf("%d", &K);
  for (int i = 0; i < K; ++i) scanf("%lld", &t[i][0]);
  scanf("%d%d", &N, &M);
  N -= K;
  for (int i = 1; i < K; ++i) t[i - 1][i] = 1;
  ans[0][0] = 1;
  ans = ans * t.Pow(N);
  a = ans[0][0];
  n = (int)ceil(sqrt(998244353));
  long long s3(Pow(3, a)), tmp(M), tt(Pow(s3, n)), res(-1);
  for (int i = 0; i <= n; ++i, tmp = (tmp * s3) % 998244353) mp[tmp] = i;
  tmp = tt;
  for (int i = 1; i <= n; ++i, tmp = (tmp * tt) % 998244353)
    if (mp.count(tmp)) return printf("%lld\n", Pow(3, i * n - mp[tmp])), 0;
  printf("-1\n");
  return 0;
}