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 = 102, P = 998244353, P2 = 998244352, G = 3;
int K;
long long b[N], n, m, C;
template <typename T>
inline void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
inline void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= f;
}
namespace BSGS {
const int S = 32000, M = 2000000;
int cnt = 0, adj[M + 5], nxt[S + 5];
long long key[S + 5], val[S + 5];
void insert(long long K, long long V) {
  int p = K % M;
  key[++cnt] = K;
  val[cnt] = V;
  nxt[cnt] = adj[p];
  adj[p] = cnt;
}
inline long long search(long long K) {
  for (int u = adj[K % M]; u; u = nxt[u])
    if (key[u] == K) return val[u];
  return -1;
}
inline void init() {
  long long sum = 1;
  for (int i = 1; i <= S; i++) sum = sum * G % P;
  long long tot = 1;
  for (int i = 1; (i - 1) * S < P - 1; i++)
    tot = tot * sum % P, insert(tot, i * S);
}
inline long long log(long long x) {
  long long sum = 1, ret;
  for (int i = 1; i <= S; i++) {
    sum = sum * G % P;
    ret = search(sum * x % P);
    if (~ret && ret < P - 1) return ret - i;
  }
  assert(0);
  return -1;
}
};  // namespace BSGS
struct Matrix {
  long long g[N][N];
  Matrix() { memset(g, 0, sizeof(g)); }
  Matrix(int x) {
    memset(g, 0, sizeof(g));
    for (int i = 1; i <= K; ++i) g[i][i] = 1;
  }
  Matrix operator*(const Matrix &b) {
    Matrix c;
    for (int i = 1; i <= K; ++i)
      for (int j = 1; j <= K; ++j)
        for (int k = 1; k <= K; ++k)
          c.g[i][j] = (c.g[i][j] + (long long)g[i][k] * b.g[k][j]) % P2;
    return c;
  }
};
inline long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    long long ret = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return ret;
  }
}
inline long long pow_mod(long long a, long long n) {
  long long b = a, res = 1;
  while (n > 0) {
    if (n & 1) res = (long long)res * b % P;
    b = (long long)b * b % P;
    n >>= 1;
  }
  return res;
}
inline Matrix qpow(Matrix a, long long n) {
  Matrix ret(1);
  while (n > 0) {
    if (n & 1) ret = ret * a;
    a = a * a;
    n >>= 1;
  }
  return ret;
}
inline long long calcC() {
  Matrix ret, op;
  ret.g[K][1] = 1;
  for (int i = 1; i < K; ++i) op.g[i][i + 1] = 1;
  for (int i = 1; i <= K; ++i) op.g[K][i] = b[K - i + 1];
  ret = qpow(op, n - K) * ret;
  return ret.g[K][1];
}
inline long long solve(long long A, long long B) {
  long long a = A, b = P2, g, x, y;
  g = exgcd(a, b, x, y);
  if (B % g) return -1;
  x *= B / g, y *= B / g;
  long long t = b / g;
  x = (x % t + t) % t;
  return x;
}
int main() {
  BSGS ::init();
  read(K);
  for (int i = 1; i <= K; ++i) read(b[i]);
  read(n), read(m);
  C = calcC();
  m = BSGS ::log(m);
  long long ans = solve(C, m);
  if (~ans)
    printf("%d\n", pow_mod(G, ans));
  else
    printf("-1\n");
  return 0;
}