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.
#include <bits/stdc++.h>
using namespace std;
const long long p = 998244353;
const long long g = 3;
int k;
long long n, m, b[210];
struct Matrix {
  int n;
  long long mat[210][210];
  Matrix() {
    n = 0;
    memset(mat, 0, sizeof(mat));
  }
  Matrix(int _n, long long diag) {
    n = _n;
    memset(mat, 0, sizeof(mat));
    for (int i = 1; i <= n; i++) {
      mat[i][i] = diag;
    }
  }
  Matrix(const Matrix &c) {
    n = c.n;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        mat[i][j] = c.mat[i][j];
      }
    }
  }
  Matrix operator*(const Matrix &a) const {
    Matrix ans = Matrix(n, 0);
    for (int k = 1; k <= n; k++) {
      for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
          ans.mat[i][j] += mat[i][k] * a.mat[k][j];
          ans.mat[i][j] %= (p - 1);
        }
      }
    }
    return ans;
  }
  Matrix mat_pow(long long t) {
    Matrix base = Matrix(*this), ans = Matrix(n, 1);
    while (t) {
      if (t & 1) {
        ans = ans * base;
      }
      base = base * base;
      t >>= 1;
    }
    return ans;
  }
};
namespace GCD {
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long ex_gcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return a;
  }
  long long q = ex_gcd(b, a % b, y, x);
  y -= a / b * x;
  return q;
}
}  // namespace GCD
namespace BSGS {
long long qpow(long long a, long long b, long long p) {
  long long ans = 1, base = a;
  while (b) {
    if (b & 1) {
      (ans *= base) %= p;
    }
    (base *= base) %= p;
    b >>= 1;
  }
  return ans;
}
long long inv(long long x, long long p) { return qpow(x, p - 2, p); }
map<long long, long long> tab;
long long bsgs(long long a, long long b, long long p) {
  long long u = (long long)sqrt(p) + 1;
  long long now = 1, step;
  for (long long i = 0; i < u; i++) {
    long long tmp = b * inv(now, p) % p;
    if (!tab.count(tmp)) {
      tab[tmp] = i;
    }
    (now *= a) %= p;
  }
  step = now;
  now = 1;
  for (long long i = 0; i < p; i += u) {
    if (tab.count(now)) {
      return i + tab[now];
    }
    (now *= step) %= p;
  }
  throw;
  return -1;
}
}  // namespace BSGS
namespace SOL {
long long solve(long long a, long long b, long long c) {
  if (c == 0) return 0;
  long long q = GCD::gcd(a, b);
  if (c % q) {
    return -1;
  }
  a /= q, b /= q, c /= q;
  long long ans, _;
  GCD::ex_gcd(a, b, ans, _);
  (ans *= c) %= b;
  while (ans < 0) ans += b;
  return ans;
}
}  // namespace SOL
int main() {
  scanf("%d", &k);
  for (int i = 1; i <= k; i++) {
    scanf("%d", &b[i]);
    b[i] %= (p - 1);
  }
  scanf("%lld%lld", &n, &m);
  Matrix A = Matrix(k, 0);
  for (int i = 1; i <= k; i++) {
    A.mat[1][i] = b[i];
  }
  for (int i = 2; i <= k; i++) {
    A.mat[i][i - 1] = 1;
  }
  A = A.mat_pow(n - k);
  long long ans = SOL::solve(A.mat[1][1], p - 1, BSGS::bsgs(g, m, p));
  if (ans >= 0) {
    printf("%lld\n", BSGS::qpow(g, ans, p));
  } else {
    puts("-1");
  }
  return 0;
}
