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 std::abs;
using std::bitset;
using std::cerr;
using std::cin;
using std::cout;
using std::map;
using std::max;
using std::min;
using std::pair;
using std::set;
using std::sort;
using std::string;
using std::swap;
using std::unordered_map;
using std::unordered_set;
using std::vector;
using ll = long long;
using uint = unsigned int;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using ull = unsigned long long;
using ld = long double;
using vci = vector<int>;
using vcl = vector<ll>;
template <typename T>
void _dbg(const char* _s, T _h) {
  cerr << _s << " = " << _h << "\n";
}
template <typename T, typename... Ts>
void _dbg(const char* _s, T _h, Ts... _t) {
  int _b = 0;
  while (((_b += *_s == '(') -= *_s == ')') != 0 || *_s != ',') cerr << *_s++;
  cerr << " = " << _h << ",";
  _dbg(_s + 1, _t...);
}
struct init {
  init() {
    cin.tie(0);
    std::iostream::sync_with_stdio(0);
    cout << std::fixed << std::setprecision(10);
    cerr << std::fixed << std::setprecision(5);
  }
  ~init() {}
} init;
template <typename T, typename U>
void upx(T& x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
void upn(T& x, U y) {
  if (x > y) x = y;
}
const ll N = 111, M = 998244353;
using mat = vector<vector<ll>>;
mat operator*(mat a, mat b) {
  ll n = a.size();
  ll m = b[0].size();
  ll kk = a[0].size();
  assert(kk == b.size());
  mat res(n, vector<ll>(m, 0));
  for (ll i = 0; i < (n); ++i)
    for (ll j = 0; j < (m); ++j) {
      for (ll k = 0; k < (kk); ++k) res[i][j] += (a[i][k] * b[k][j]) % (M - 1);
      res[i][j] %= (M - 1);
    }
  return res;
}
mat binpow(mat a, ll pw) {
  ll n = a.size();
  mat res(n, vector<ll>(n, 0));
  for (ll i = 0; i < (n); ++i) res[i][i] = 1;
  while (pw) {
    if (pw & 1) {
      res = res * a;
    }
    a = a * a;
    pw /= 2;
  }
  return res;
}
ll gcd(ll a, ll b) { return a ? gcd(b % a, a) : b; }
ll powmod(ll a, ll b, ll p) {
  ll res = 1;
  while (b)
    if (b & 1)
      res = ll(res * 1ll * a % p), --b;
    else
      a = ll(a * 1ll * a % p), b >>= 1;
  return res;
}
ll generator(ll p) {
  vector<ll> fact;
  ll phi = p - 1, n = phi;
  for (ll 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 (ll 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;
}
ll discroot(ll n, ll k, ll a) {
  if (a == 0) {
    return 1;
  }
  ll g = generator(n);
  ;
  ll sq = (ll)sqrt(n + .0) + 1;
  std::unordered_map<ll, ll> dec;
  for (ll i = 1; i <= sq; ++i)
    dec[powmod(g, ll(i * sq * 1ll * k % (n - 1)), n)] = i;
  ;
  ll any_ans = -1;
  ;
  for (ll i = 0; i < sq; ++i) {
    ll my = ll(powmod(g, ll(i * 1ll * k % (n - 1)), n) * 1ll * a % n);
    auto it = dec.find(my);
    if (it != dec.end()) {
      any_ans = it->second * sq - i;
      break;
    }
  };
  if (any_ans == -1) {
    return -1;
  };
  ll delta = (n - 1) / gcd(k, n - 1);
  vector<ll> ans;
  for (ll cur = any_ans % delta; cur < n - 1; cur += delta) {
    return powmod(g, cur, n);
  }
}
ll b[N];
int32_t main() {
  ll n, m, k;
  cin >> k;
  for (ll i = 0; i < (k); ++i) cin >> b[i];
  cin >> n >> m;
  mat a(k, vector<ll>(k, 0));
  for (ll i = 0; i < (k); ++i) a[0][i] = b[i];
  for (ll i = 0; i < (k - 1); ++i) a[i + 1][i] = 1;
  a = binpow(a, n - k);
  ll a11 = a[0][0];
  ;
  cout << discroot(M, a11, m) << '\n';
  return 0;
}