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>
#pragma comment(linker, "/stack:247474112")
#pragma GCC optimize("Ofast")
using namespace std;
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
const int mod = 998244353;
long long linearRec(vector<long long> S, vector<long long> tr, long long k) {
  long long n = S.size();
  auto combine = [&](vector<long long> a, vector<long long> b) {
    vector<long long> res(n * 2 + 1);
    for (long long i = 0; i < n + 1; i++)
      for (long long j = 0; j < n + 1; j++)
        res[i + j] = (res[i + j] + a[i] * b[j]) % (mod - 1);
    for (long long i = 2 * n; i > n; --i)
      for (long long j = 0; j < n; j++)
        res[i - 1 - j] = (res[i - 1 - j] + res[i] * tr[j]) % (mod - 1);
    res.resize(n + 1);
    return res;
  };
  vector<long long> pol(n + 1), e(pol);
  pol[0] = e[1] = 1;
  for (++k; k; k /= 2) {
    if (k % 2) pol = combine(pol, e);
    e = combine(e, e);
  }
  long long res = 0;
  for (long long i = 0; i < n; i++) res = (res + pol[i + 1] * S[i]) % mod;
  return res;
}
long long k, n, m;
vector<long long> b;
int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
int powmod(int a, int b, int p) {
  int res = 1;
  while (b)
    if (b & 1)
      res = int(res * 1ll * a % p), --b;
    else
      a = int(a * 1ll * a % p), b >>= 1;
  return res;
}
int generator(int p) {
  vector<int> fact;
  int phi = p - 1, n = phi;
  for (int i = 2; i * i <= n; ++i)
    if (n % i == 0) {
      fact.push_back(i);
      while (n % i == 0) n /= i;
    }
  if (n > 1) fact.push_back(n);
  for (int res = 2; res <= p; ++res) {
    bool ok = true;
    for (size_t i = 0; i < fact.size() && ok; ++i)
      ok &= powmod(res, phi / fact[i], p) != 1;
    if (ok) return res;
  }
  return -1;
}
int discrete_root(int n, int k, int a) {
  if (a == 0) return 0;
  int g = generator(n);
  int sq = (int)sqrt(n + .0) + 1;
  vector<pair<int, int> > dec(sq);
  for (int i = 1; i <= sq; ++i)
    dec[i - 1] = make_pair(powmod(g, int(i * sq * 1ll * k % (n - 1)), n), i);
  sort(dec.begin(), dec.end());
  int any_ans = -1;
  for (int i = 0; i < sq; ++i) {
    int my = int(powmod(g, int(i * 1ll * k % (n - 1)), n) * 1ll * a % n);
    vector<pair<int, int> >::iterator it =
        lower_bound(dec.begin(), dec.end(), make_pair(my, 0));
    if (it != dec.end() && it->first == my) {
      any_ans = it->second * sq - i;
      break;
    }
  }
  if (any_ans == -1) return -1;
  int delta = (n - 1) / gcd(k, n - 1);
  vector<int> ans;
  for (int cur = any_ans % delta; cur < n - 1; cur += delta)
    return powmod(g, cur, n);
}
void Input() {
  cin >> k;
  b.resize(k);
  for (auto& z : b) cin >> z;
  cin >> n >> m;
}
void Solve() {
  vector<long long> series(k, 0);
  series[k - 1] = 1;
  long long expo = linearRec(series, b, n - 1) % (mod - 1);
  cout << discrete_root(mod, expo, m) << '\n';
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  Input();
  Solve();
  return 0;
}