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 mpow(long long a, long long b, long long mod) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
int mlog(long long r, long long b, long long mod) {
  map<long long, int> mp;
  for (int i = 99999; i >= 0; i--) {
    mp[mpow(r, i, mod)] = i;
  }
  for (int i = 0; i < mod - 1; i += 100000) {
    if (mp.count(b * mpow(r, mod - 1 - i, mod) % mod)) {
      return i + mp[b * mpow(r, mod - 1 - i, mod) % mod];
    }
  }
  return -1;
}
vector<long long> pmod(vector<long long> a, vector<long long> p,
                       long long mod) {
  const int n = a.size();
  const int m = p.size();
  for (int i = n - 1; i >= m; i--) {
    for (int j = 0; j < m; j++) {
      a[i - m + j] += a[i] * p[j];
      a[i - m + j] %= mod;
    }
  }
  a.resize(m);
  return a;
}
vector<long long> pmul(vector<long long> a, vector<long long> b,
                       vector<long long> p, long long mod) {
  const int n = a.size();
  const int m = b.size();
  vector<long long> c(n + m - 1);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      c[i + j] += a[i] * b[j];
      c[i + j] %= mod;
    }
  }
  return pmod(c, p, mod);
}
vector<long long> ppow(vector<long long> a, long long b, vector<long long> p,
                       long long mod) {
  vector<long long> res(1, 1);
  while (b > 0) {
    if (b & 1) res = pmul(res, a, p, mod);
    a = pmul(a, a, p, mod);
    b >>= 1;
  }
  return res;
}
long long extgcd(long long a, long long b, long long c, long long mod) {
  long long s = 1;
  long long t = 0;
  while (b != 0) {
    long long q = a / b;
    a -= b * q;
    s -= t * q;
    swap(a, b);
    swap(s, t);
  }
  if (c % a != 0) return -1;
  if (s < 0) s += mod;
  return c / a * s % mod;
}
int main() {
  int K;
  long long N, M;
  scanf("%d", &K);
  vector<long long> p(K);
  for (int i = K - 1; i >= 0; i--) {
    scanf("%lld", &p[i]);
    p[i] %= 998244353 - 1;
  }
  scanf("%lld %lld", &N, &M);
  N--;
  M = mlog(3, M, 998244353);
  vector<long long> a(2);
  a[1] = 1;
  a = ppow(a, N, p, 998244353 - 1);
  long long ans = extgcd(a[K - 1], 998244353 - 1, M, 998244353 - 1);
  if (ans == -1) {
    puts("-1");
  } else {
    ans = mpow(3, ans, 998244353);
    printf("%lld\n", ans);
  }
  return 0;
}