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 maxn = 1e2 + 5;
const int inf = 0x3f3f3f3f;
long long mod = 998244353;
struct Mat {
  const static int N = 1e2 + 5;
  long long a[N][N];
  int n;
  Mat(int n) : n(n) { memset(a, 0, sizeof a); }
  void setI() {
    memset(a, 0, sizeof a);
    for (int i = 0; i < n; ++i) a[i][i] = 1;
  }
  Mat operator*(const Mat &o) const {
    Mat ret(n);
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        for (int k = 0; k < n; ++k)
          ret.a[i][j] = (ret.a[i][j] + a[i][k] * o.a[k][j]) % mod;
    return ret;
  }
};
Mat qpow(Mat a, long long b) {
  Mat ret(a.n);
  ret.setI();
  while (b) {
    if (b & 1) ret = ret * a;
    a = a * a;
    b >>= 1;
  }
  return ret;
}
long long qpow(long long a, long long b, long long p) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return ret;
}
void exgcd(long long a, long long b, long long &x, long long &y, long long &d) {
  if (!b)
    x = 1, y = 0, d = a;
  else
    exgcd(b, a % b, y, x, d), y -= a / b * x;
}
long long get_gk(long long a, long long b, long long c) {
  long long x, y, d;
  exgcd(a, b, x, y, d);
  if (c % d) return -1;
  x *= c / d;
  long long tmp = b / d;
  x = (x % tmp + tmp) % tmp;
  return x;
}
long long inv(long long a, long long p) { return qpow(a, p - 2, p); }
map<int, int> mp;
long long BSGS(long long a, long long b, long long p) {
  a %= p, b %= p;
  if (!a && b) return -1;
  mp.clear();
  long long lim = ceil(sqrt(p));
  long long cur = 1;
  for (int i = 0; i < lim; ++i) {
    if (!mp.count(cur)) mp[cur] = i;
    cur = cur * a % p;
  }
  long long inva = inv(cur, p);
  for (int i = 0; i <= lim; ++i) {
    if (mp.count(b)) return i * lim + mp[b];
    b = b * inva % p;
  }
  return -1;
}
long long b[maxn];
long long n, k, m;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> k;
  for (int i = 1; i <= k; ++i) cin >> b[i];
  cin >> n >> m;
  Mat A(k);
  for (int i = 0; i < k; ++i) A.a[0][i] = b[i + 1];
  for (int i = 1; i < k; ++i) A.a[i][i - 1] = 1;
  --mod;
  A = qpow(A, n - k);
  ++mod;
  long long g_n = BSGS(3, m, mod);
  --mod;
  long long g_k = get_gk(A.a[0][0], mod, g_n);
  if (g_k == -1) {
    cout << "-1" << endl;
    return 0;
  }
  ++mod;
  long long f_k = qpow(3, g_k, mod);
  cout << f_k << endl;
  return 0;
}