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 int N = 101;
long long n, arr[N];
long long pwr(int b, int p, int mod) {
  if (!p) return 1;
  long long ans = pwr(b, p >> 1, mod);
  ans = ans * ans % mod;
  if (p & 1) ans = ans * b % mod;
  return ans;
}
struct matrix {
  int n;
  int M[N][N];
  matrix() {}
  matrix(int _n) {
    n = _n;
    memset(M, 0, sizeof M);
  }
  matrix operator*(const matrix& mat) const {
    matrix ret(n);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < n; k++) {
          ret.M[i][j] =
              (ret.M[i][j] + 1LL * M[i][k] * mat.M[k][j] % (998244353 - 1)) %
              (998244353 - 1);
        }
      }
    }
    return ret;
  }
};
matrix I;
matrix matrixPw(matrix& m, int p, int mod) {
  if (!p) return I;
  matrix ans = matrixPw(m, p >> 1, mod);
  ans = ans * ans;
  if (p & 1) ans = ans * m;
  return ans;
}
int generator(int p) {
  vector<int> fact;
  int phi = p - 1, n = phi;
  for (int 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 (int res = 2; res <= p; ++res) {
    bool ok = true;
    for (size_t i = 0; i < fact.size() && ok; ++i)
      ok &= pwr(res, phi / fact[i], p) != 1;
    if (ok) return res;
  }
  return -1;
}
long long baby_giant(int a, int b, int m) {
  int n = (int)sqrt(m + .0) + 1;
  map<int, int> vals;
  for (int p = n; p >= 1; --p) vals[pwr(a, p * n, m)] = p;
  for (int q = 0; q <= n; ++q) {
    int cur = (pwr(a, q, m) * b) % m;
    if (vals.count(cur)) {
      long long ans = 1LL * vals[cur] * n - q;
      return ans;
    }
  }
  return -1;
}
int main() {
  int k;
  scanf("%d", &k);
  vector<int> vec(k);
  for (int i = 0; i < k; i++) scanf("%d", &vec[i]);
  int n, m;
  scanf("%d%d", &n, &m);
  I = matrix(k);
  for (int i = 0; i < k; i++) I.M[i][i] = 1;
  matrix T(k);
  for (int i = 0; i < k - 1; i++) T.M[i][i + 1] = 1;
  for (int i = 0; i < k; i++) T.M[k - 1][k - 1 - i] = vec[i];
  matrix TT = matrixPw(T, n - k, 998244353 - 1);
  long long pw = TT.M[k - 1][k - 1];
  if (true) {
    int g = generator(998244353);
    int y = baby_giant(pwr(g, pw, 998244353), m, 998244353);
    if (y == -1) return printf("-1"), 0;
    printf("%lld", pwr(g, y, 998244353));
  } else {
    int y = pwr(m, pwr(pw, 402653183, 998244353 - 1), 998244353);
    printf("%d", y);
  }
  return 0;
}