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 mod = 998244353;
struct Matrix {
  long long m[100][100];
  void init() {
    memset(m, 0, sizeof(m));
    for (int i = 0; i < 100; i++) m[i][i] = 1;
  }
  Matrix operator*(const Matrix &a) const {
    Matrix ret;
    memset(ret.m, 0, sizeof(ret.m));
    for (int i = 0; i < 100; i++) {
      for (int j = 0; j < 100; j++) {
        for (int k = 0; k < 100; k++) {
          ret.m[i][j] =
              (ret.m[i][j] + m[i][k] * a.m[k][j] % (mod - 1)) % (mod - 1);
        }
      }
    }
    return ret;
  }
};
Matrix quick(Matrix a, long long b) {
  Matrix ret;
  ret.init();
  while (b) {
    if (b & 1) ret = ret * a;
    a = a * a;
    b >>= 1;
  }
  return ret;
}
map<long long, long long> mp;
long long pow(long long a, long long b, long long p) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % p;
    b >>= 1;
    a = a * a % p;
  }
  return ans;
}
long long BSGS(long long A, long long B, long long C) {
  mp.clear();
  if (A % C == 0) return -2;
  long long m = ceil(sqrt(C));
  long long ans;
  for (int i = 0; i <= m; i++) {
    if (i == 0) {
      ans = B % C;
      mp[ans] = i;
      continue;
    }
    ans = (ans * A) % C;
    mp[ans] = i;
  }
  long long t = pow(A, m, C);
  ans = t;
  for (int i = 1; i <= m; i++) {
    if (i != 1) ans = ans * t % C;
    if (mp.count(ans)) {
      int ret = i * m % C - mp[ans] % C;
      return (ret % C + C) % C;
    }
  }
  return -2;
}
long long extend_gcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0 && b == 0) return -1;
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extend_gcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
long long brr[105];
Matrix base;
int main() {
  int k;
  scanf("%d", &k);
  for (int i = 0; i < k; i++) {
    scanf("%lld", &brr[i]);
    base.m[0][i] = brr[i];
    if (i != k - 1) base.m[i + 1][i] = 1;
  }
  int n, m;
  scanf("%d%d", &n, &m);
  Matrix A = quick(base, n - k);
  long long hn = BSGS(3, m, mod);
  long long a = A.m[0][0];
  long long c = 0;
  for (int i = 1; i < k; i++) c = (c + A.m[0][i]) % (mod - 1);
  long long p = mod - 1;
  long long x, y;
  long long gcd = extend_gcd(a, p, x, y);
  if (hn % gcd != 0)
    puts("-1");
  else {
    x += p;
    x %= p;
    x = x * (hn / gcd) % p;
    printf("%lld\n", pow(3, x, mod));
  }
}