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;
int mul(int a, int b, int p) {
  long long res = (long long)a * b;
  return res % p;
}
int add(int a, int b, int p) {
  long long res = (long long)a + b;
  return res % p;
}
struct Matrix {
  int n, m;
  vector<vector<int>> a;
  Matrix() {
    n = -1;
    m = -1;
  }
  Matrix(int n_, int m_) {
    n = n_;
    m = m_;
    a.resize(n, vector<int>(m));
  }
  void set(int i, int j, int val) { a[i][j] = val; }
  friend Matrix mul(const Matrix &f, const Matrix &s, int p) {
    int n = (int)f.a.size();
    int m = (int)f.a[0].size();
    int k = (int)s.a[0].size();
    Matrix res(n, m);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        int val = 0;
        for (int t = 0; t < k; t++) {
          val = add(val, mul(f.a[i][t], s.a[t][j], p), p);
        }
        res.set(i, j, val);
      }
    }
    return res;
  }
};
Matrix binpow(Matrix f, int b, int p) {
  int n = (int)f.a.size();
  Matrix res(n, n);
  for (int i = 0; i < n; i++) {
    res.set(i, i, 1);
  }
  while (b > 0) {
    if (b % 2 == 1) {
      res = mul(res, f, p);
      b--;
    } else {
      f = mul(f, f, p);
      b /= 2;
    }
  }
  return res;
}
int binpow(int f, int b, int p) {
  int res = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      res = mul(res, f, p);
      b--;
    } else {
      f = mul(f, f, p);
      b /= 2;
    }
  }
  return res;
}
int solve_discrete_logging(int a, int b, int p) {
  const int LIM = 32000;
  map<int, vector<int>> can;
  for (int i = 0; i < LIM; i++) {
    auto cur = binpow(a, i, p);
    can[cur].push_back(i);
  }
  for (int i = 0; i < LIM; i++) {
    int inv = binpow(b, p - 2, p);
    int cur = binpow(a, i * LIM, p);
    cur = mul(cur, inv, p);
    int need = binpow(cur, p - 2, p);
    if (can.count(need)) {
      cerr << "answer " << i * LIM + can[need].front() << endl;
      return i * LIM + can[need].front();
    }
  }
  return -1;
}
int solve(int a, int b, int p) {
  int g = 3;
  auto t = binpow(g, a, p);
  int y = solve_discrete_logging(t, b, p);
  if (y == -1) {
    return -1;
  }
  int x = binpow(g, y, p);
  return x;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  const int mod = 998244353;
  int k;
  cin >> k;
  Matrix base(k, 1);
  base.set(0, 0, 1);
  vector<int> b(k);
  for (auto &t : b) {
    cin >> t;
  }
  Matrix go(k, k);
  for (int j = 0; j < k; j++) {
    go.set(0, j, b[j]);
  }
  for (int i = 1; i < k; i++) {
    go.set(i, i - 1, 1);
  }
  int n, m;
  cin >> n >> m;
  int to = n - (k);
  Matrix pw = binpow(go, to, mod - 1);
  Matrix res = mul(pw, base, mod - 1);
  int pow = res.a[0][0];
  int ans = solve(pow, m, mod);
  cout << ans << endl;
}