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 GCC optimize("O3")
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template <class T>
using V = vector<T>;
const ll inf = (1e18);
const vector<int> dy = {-1, 0, 1, 0}, dx = {0, -1, 0, 1};
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
ll LCM(ll c, ll d) { return c / GCD(c, d) * d; }
struct __INIT {
  __INIT() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
  }
} __init;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
void debag(const vector<T> &a) {
  cerr << "debag :";
  for (auto v : a) cerr << v << " ";
  cerr << "\n";
}
template <class T>
void print(const vector<T> &a) {
  for (auto v : a) cout << v << " ";
  cout << "\n";
}
ll mod = 998244352;
struct mint {
  using ull = unsigned long long int;
  ull v;
  mint(ll vv = 0) { s(vv % mod + mod); }
  mint &s(ull vv) {
    v = vv < mod ? vv : vv - mod;
    return *this;
  }
  mint operator-() const { return mint() - *this; }
  mint &operator+=(const mint &val) { return s(v + val.v); }
  mint &operator-=(const mint &val) { return s(v + mod - val.v); }
  mint &operator*=(const mint &val) {
    v = ull(v) * val.v % mod;
    return *this;
  }
  mint &operator/=(const mint &val) { return *this *= val.inv(); }
  mint operator+(const mint &val) { return mint(*this) += val; }
  mint operator-(const mint &val) { return mint(*this) -= val; }
  mint operator*(const mint &val) { return mint(*this) *= val; }
  mint operator/(const mint &val) { return mint(*this) /= val; }
  mint pow(ll n) const {
    mint res(1), x(*this);
    while (n) {
      if (n & 1) res *= x;
      x *= x;
      n >>= 1ll;
    }
    return res;
  }
  mint inv() const { return pow(mod - 2); }
  friend ostream &operator<<(ostream &os, const mint &val) {
    return os << val.v;
  }
  bool operator<(const mint &val) const { return v < val.v; }
  bool operator==(const mint &val) const { return v == val.v; }
  bool operator>(const mint &val) const { return v > val.v; }
};
template <class T>
struct Matrix {
  vector<vector<T>> A;
  Matrix() {}
  Matrix(size_t n, size_t m) : A(n, vector<T>(m, 0)) {}
  Matrix(size_t n) : A(n, vector<T>(n, 0)) {}
  size_t height() const { return (A.size()); }
  size_t width() const { return (A[0].size()); }
  inline const vector<T> &operator[](int k) const { return (A.at(k)); }
  inline vector<T> &operator[](int k) { return (A.at(k)); }
  static Matrix I(size_t n) {
    Matrix mat(n);
    for (int i = 0; i < n; i++) mat[i][i] = (T)(1);
    return mat;
  }
  Matrix &operator+=(const Matrix &B) {
    size_t n = height(), m = width();
    assert(n == B.height() && m == B.width());
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        (*this)[i][j] += B[i][j];
      }
    }
    return (*this);
  }
  Matrix &operator-=(const Matrix &B) {
    size_t n = height(), m = width();
    assert(n == B.height() && m == B.width());
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        (*this)[i][j] -= B[i][j];
      }
    }
    return (*this);
  }
  Matrix &operator*=(const Matrix &B) {
    size_t n = height(), m = B.width(), w = height();
    assert(w == B.height());
    vector<vector<T>> C(n, vector<T>(m, 0));
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        for (int k = 0; k < w; k++) {
          C[i][j] += (*this)[i][k] * B[k][j];
        }
      }
    }
    A.swap(C);
    return (*this);
  }
  Matrix &operator^=(long long k) {
    Matrix B = Matrix::I(height());
    while (k > 0) {
      if (k & 1) {
        B *= (*this);
      }
      (*this) *= (*this);
      k >>= 1ll;
    }
    A.swap(B.A);
    return (*this);
  }
  Matrix operator+(const Matrix &B) const { return (Matrix(*this) += B); }
  Matrix operator-(const Matrix &B) const { return (Matrix(*this) -= B); }
  Matrix operator*(const Matrix &B) const { return (Matrix(*this) *= B); }
  Matrix operator^(const long long k) const { return (Matrix(*this) ^= k); }
  friend ostream &operator<<(ostream &os, Matrix &mat) {
    size_t n = mat.height(), m = mat.width();
    for (int i = 0; i < n; i++) {
      os << "[";
      for (int j = 0; j < m; j++) {
        os << mat[i][j] << (j + 1 == m ? "]\n" : ",");
      }
    }
    return (os);
  }
  T det() {
    Matrix B(*this);
    assert(height() == width());
    T ret = (T)(1);
    for (int i = 0; i < width(); i++) {
      int idx = -1;
      for (int j = i; j < width(); j++) {
        if (B[j][i] != 0) idx = j;
      }
      if (idx == -1) return (T)(0);
      if (i != idx) {
        ret *= (T)(-1);
        swap(B[i], B[idx]);
      }
      ret *= B[i][i];
      T v = B[i][i];
      for (int j = 0; j < width(); j++) {
        B[i][j] /= v;
      }
      for (int j = i + 1; j < width(); j++) {
        T a = B[j][i];
        for (int k = 0; k < width(); k++) {
          B[j][k] -= B[i][k] * a;
        }
      }
    }
    return (ret);
  }
};
ll babystep_giantstep(mint A, mint B) {
  mint tmp = mint(1);
  map<ll, ll> mp;
  ll sq = ll(sqrt(mod)) + 1;
  for (ll i = 1; i <= sq; i++) {
    tmp *= A;
    if (!mp.count(tmp.v)) mp[tmp.v] = i;
  }
  mint r = tmp.inv(), cp = B;
  for (ll i = 0; i <= sq; i++) {
    if (mp.count(cp.v)) {
      return mp[cp.v] + i * sq;
    }
    cp *= r;
  }
  return -1;
}
int main() {
  int n;
  cin >> n;
  V<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  Matrix<mint> res(n, 1), rec(n);
  for (int i = 0; i < n; i++) {
    rec[0][i] = mint(a[i]);
  }
  for (int i = 1; i < n; i++) {
    rec[i][i - 1] = 1;
  }
  res[0][0] = 1;
  ll v, val;
  cin >> v >> val;
  rec ^= (v - n);
  rec *= res;
  ll k = rec[0][0].v;
  mod++;
  ll ans = babystep_giantstep(mint(3).pow(k), mint(val));
  if (ans == -1)
    cout << -1 << "\n";
  else
    cout << mint(3).pow(ans) << "\n";
}