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;
long long n, m, f;
struct M {
  long long t[110][110] = {{0}};
};
M mul(M A, M B) {
  M ans;
  long long i, j, k;
  for (k = 0; k < m; k++)
    for (i = 0; i < m; i++)
      for (j = 0; j < m; j++)
        ans.t[i][j] = (ans.t[i][j] + A.t[i][k] * B.t[k][j]) % (998244353 - 1);
  return ans;
}
M qpow(M A, long long x) {
  M ans;
  long long i;
  for (i = 0; i < m; i++) ans.t[i][i] = 1;
  while (x) {
    if (x & 1LL) ans = mul(ans, A);
    A = mul(A, A);
    x >>= 1LL;
  }
  return ans;
}
long long qpow(long long a, long long x) {
  long long ans = 1;
  while (x) {
    if (x & 1LL) ans = ans * a % 998244353;
    a = a * a % 998244353;
    x >>= 1LL;
  }
  return ans;
}
map<long long, long long> mp;
map<long long, long long>::iterator ite;
long long bsgs(long long a, long long t) {
  long long i, x = 1, y, k = ceil(pow(998244353, 0.5));
  for (i = 0; i <= k; i++) {
    mp[t * x % 998244353] = i;
    x = x * a % 998244353;
  }
  x = 1;
  y = qpow(a, k);
  for (i = 0; i < k; i++) {
    ite = mp.find(x);
    if (ite != mp.end()) return ((k * i - ite->second) + 998244353) % 998244353;
    x = x * y % 998244353;
  }
  return -1;
}
void extgcd(long long a, long long b, long long &g, long long &x,
            long long &y) {
  if (b == 0) {
    g = a;
    x = 1;
    y = 0;
    return;
  }
  extgcd(b, a % b, g, y, x);
  y -= a / b * x;
}
long long solve(long long a, long long t) {
  long long rhs = bsgs(3, t), g, x, y;
  extgcd(a, 998244353 - 1, g, x, y);
  if (rhs % g == 0) {
    x = x * (rhs / g) % (998244353 - 1);
    x = (x + 998244353 - 1) % (998244353 - 1);
    return qpow(3, x);
  } else
    return -1;
}
int main() {
  M A;
  long long i;
  scanf("%lld", &m);
  for (i = 0; i < m; i++) scanf("%lld", &A.t[0][i]);
  for (i = 1; i < m; i++) A.t[i][i - 1] = 1;
  scanf("%lld%lld", &n, &f);
  A = qpow(A, n - m);
  printf("%lld", solve(A.t[0][0], f));
}