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 MAX_MAT = 150, mod = 998244353;
int n, m, k, ki[MAX_MAT];
struct matrix {
  long long mat[MAX_MAT][MAX_MAT];
  void basify() {
    for (int i = 1; i <= k; i++) mat[i][i] = 1;
  }
  long long *operator[](const int &id) { return mat[id]; }
  matrix operator*(const matrix &mt) const {
    matrix ans;
    memset(ans.mat, 0, sizeof(ans.mat));
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= k; j++)
        for (int mid = 1; mid <= k; mid++)
          ans.mat[i][j] =
              (ans.mat[i][j] + mat[i][mid] * mt.mat[mid][j] % (mod - 1)) %
              (mod - 1);
    return ans;
  }
  matrix operator^(const int &tim) const {
    matrix ans, bas = *this;
    ans.basify();
    int ti = tim;
    while (ti) {
      if (ti & 1) ans = ans * bas;
      bas = bas * bas;
      ti >>= 1;
    }
    return ans;
  }
};
long long quick_pow(long long bas, long long tim) {
  long long ans = 1;
  while (tim) {
    if (tim & 1) ans = ans * bas % mod;
    bas = bas * bas % mod;
    tim >>= 1;
  }
  return ans;
}
long long bsgs(long long a, long long y) {
  if (a == 0 && y == 0) return 1;
  if (a == 0 && y != 0) return -1;
  map<long long, long long> hsh;
  long long u = ceil(sqrt(mod));
  for (int i = 0, x = 1; i <= u; i++, x = x * a % mod) hsh[y * x % mod] = i;
  long long unit = quick_pow(a, u);
  for (int i = 1, x = unit; i <= u; i++, x = x * unit % mod)
    if (hsh.count(x)) return i * u - hsh[x];
  return -1;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, x, y), z = x;
  x = y, y = z - (a / b) * y;
  return d;
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long solve(long long a, long long b, long long c) {
  if (b == 0) return 0;
  long long d = gcd(a, b);
  a /= d, b /= d;
  if (gcd(a, c) != 1) return -1;
  long long res, tmp;
  exgcd(a, c, res, tmp);
  res = res * b % c;
  res = (res + c) % c;
  return res;
}
int main() {
  scanf("%d", &k);
  for (int i = 1; i <= k; i++) scanf("%d", &ki[i]);
  scanf("%d%d", &n, &m);
  matrix B;
  for (int i = 1; i <= k; i++) B[1][i] = ki[i];
  for (int i = 2; i <= k; i++) B[i][i - 1] = 1;
  B = B ^ (n - k);
  long long res = B[1][1], s = bsgs(3, m), ans = solve(res, s, mod - 1);
  if (ans == -1)
    puts("-1");
  else
    printf("%lld", quick_pow(3, ans));
  return 0;
}