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;
const long long MOD = 998244353;
const long long K = 4e4;
const long long G = 3;
long long add(long long a, long long b) { return (a + b) % MOD; }
long long sub(long long a, long long b) { return (a + MOD - b) % MOD; }
long long mul(long long a, long long b) { return (long long)(a * b % MOD); }
long long power_mod(long long a, long long p) {
  long long x = 1;
  for (; p > 0; p /= 2) {
    if (p % 2) {
      x = mul(x, a);
    }
    a = mul(a, a);
  }
  return x;
}
long long divide(long long a, long long b) {
  return mul(a, power_mod(b, MOD - 2));
}
long long discrete_log(long long x) {
  unordered_map<long long, long long> m;
  for (long long i = 0; i < K; ++i) {
    long long r = power_mod(G, mul(i, K));
    if (!m.count(r)) {
      m[r] = i;
    }
  }
  for (long long j = 0; j < K; ++j) {
    long long r = mul(x, power_mod(G, j));
    if (m.count(r)) {
      return sub(mul(m[r], K), j);
    }
  }
  return -1;
}
using Scalar = long long;
struct Matrix {
  long long n, m;
  vector<Scalar> data;
  Matrix() : n(0), m(0), data(0) {}
  Matrix(long long x, long long y) : n(x), m(y), data(x * y, 0) {}
  Scalar &operator()(long long i, long long j) { return data[i * n + j]; }
  Scalar operator()(long long i, long long j) const { return data[i * n + j]; }
};
Matrix identity(long long n) {
  Matrix a(n, n);
  for (long long i = 0; i < n; ++i) {
    a(i, i) = 1;
  }
  return a;
}
Matrix transpose(const Matrix &a) {
  long long n = a.n, m = a.m;
  Matrix b(m, n);
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < m; ++j) {
      b(i, j) = a(j, i);
    }
  }
  return b;
}
long long new_mul(long long a, long long b) {
  return (long long)(a * b % (MOD - 1));
}
long long new_add(long long a, long long b) { return (a + b) % (MOD - 1); }
Matrix operator*(const Matrix &a, Matrix b) {
  assert(a.m == b.n);
  long long n = a.n, m = b.m, k = a.m;
  b = transpose(b);
  Matrix c(n, m);
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < m; ++j) {
      for (long long t = 0; t < k; ++t) {
        c(i, j) = new_add(c(i, j), new_mul(a(i, t), b(j, t)));
      }
    }
  }
  return c;
}
Matrix fast_power(Matrix a, long long p) {
  assert(a.n == a.m);
  long long n = a.n;
  Matrix x = identity(n);
  for (; p > 0; p >>= 1) {
    if (p & 1) {
      x = x * a;
    }
    a = a * a;
  }
  return x;
}
ostream &operator<<(ostream &out, const Matrix &a) {
  for (long long i = 0; i < a.n; ++i) {
    for (long long j = 0; j < a.m; ++j) {
      out << a(i, j) << '\t';
    }
    out << '\n';
  }
  return out;
}
long long ex_gcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long g = ex_gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return g;
}
bool diofant(long long a, long long &x, long long b, long long &y,
             long long c) {
  long long g = ex_gcd(a, b, x, y);
  if (c % g) {
    return false;
  }
  x *= c / g;
  y *= c / g;
  assert(a * x + b * y == c);
  return true;
}
long long solve(long long a, long long b) {
  long long ans, _;
  if (!diofant(a, ans, MOD - 1, _, b)) {
    return -1;
  } else {
    return (long long)((ans % (MOD - 1) + (MOD - 1)) % (MOD - 1));
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long k;
  cin >> k;
  Matrix m(k, k);
  for (long long i = 0; i < k; ++i) {
    cin >> m(0, i);
  }
  for (long long i = 1; i < k; ++i) {
    m(i, i - 1) = 1;
  }
  long long n, v;
  cin >> n >> v;
  v = discrete_log(v);
  m = fast_power(m, n - k);
  long long answer = solve(m(0, 0), v);
  if (answer < 0) {
    cout << -1 << '\n';
  } else {
    cout << power_mod(G, answer) << '\n';
  }
  return 0;
}