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;
long long gcd(long long a, long long b) {
  if (a > b) {
    swap(a, b);
  }
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
long long mp(long long a, long long n, long long m) {
  long long r = 1;
  while (n) {
    if (n & 1) {
      r = r * a % m;
    }
    a = a * a % m;
    n >>= 1;
  }
  return r;
}
long long BG(long long X, long long Y, long long M) {
  unordered_map<long long, long long> D;
  D[1] = 1;
  long long S = 31596;
  long long Z = 1;
  for (int i = 1; i <= S; i++) {
    Z = (Z * X) % M;
    D[Z] = i + 1;
  }
  if (D[Y]) {
    return D[Y] - 1;
  }
  long long R = mp(Z, M - 2, M);
  for (int i = 1; i <= S; i++) {
    Y = Y * R % M;
    if (D[Y]) {
      return D[Y] + i * S - 1;
    }
  }
  return -1;
}
const long long mod = 998244352;
vector<vector<long long>> m(vector<vector<long long>> a,
                            vector<vector<long long>> b, long long md) {
  vector<vector<long long>> r(a.size(), vector<long long>(b[0].size(), 0));
  for (int i = 0; i < (int)(a.size()); i++) {
    for (int k = 0; k < (int)(b.size()); k++) {
      for (int j = 0; j < (int)(b[0].size()); j++) {
        r[i][j] = (r[i][j] + a[i][k] * b[k][j]) % md;
      }
    }
  }
  return r;
}
vector<vector<long long>> p(vector<vector<long long>> a, long long n,
                            long long md) {
  vector<vector<long long>> r(a.size(), vector<long long>(a.size(), 0));
  for (int i = 0; i < (int)(a.size()); i++) {
    r[i][i] = 1;
  }
  while (n) {
    if (n & 1) {
      r = m(a, r, md);
    }
    a = m(a, a, md);
    n >>= 1;
  }
  return r;
}
int main() {
  long long K, N, M;
  scanf("%lld", &K);
  vector<long long> B(K);
  for (int i = 0; i < K; i++) {
    scanf("%lld", &B[i]);
  }
  scanf("%lld%lld", &N, &M);
  if (M == 1) {
    printf("1\n");
    return 0;
  }
  vector<vector<long long>> X(K, vector<long long>(K, 0));
  vector<vector<long long>> Y(K, vector<long long>(K, 0));
  Y[0][K - 1] = 1;
  for (int i = 0; i < K - 1; i++) {
    X[i + 1][i] = 1;
    X[0][i] = (B[i] % mod);
  }
  X[0][K - 1] = (B[K - 1] % mod);
  vector<vector<long long>> Z = m(p(X, N - K, mod), Y, mod);
  long long W = Z[0][K - 1];
  long long V = BG(3, M, mod + 1);
  if (W == 0) {
    printf("-1\n");
    return 0;
  }
  long long G = gcd(V, mod), H = gcd(W, mod);
  if (G % H) {
    printf("-1\n");
    return 0;
  }
  long long a = W / H, b = mod / H, u = 1, v = 0;
  while (b) {
    long long t = a / b;
    u -= t * v;
    swap(u, v);
    a -= t * b;
    swap(a, b);
  }
  u %= (mod / H);
  if (u < 0) {
    u += (mod / H);
  }
  printf("%lld\n", mp(3, (u * (V / H)) % (mod / H), mod + 1));
  return 0;
}