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 long long mod = 998244353;
struct matrix {
  int len;
  long long mat[101][101];
  matrix(int l = 0, int type = 0) {
    len = l;
    if (type == 0) {
      for (int i = 0; i < len; i++)
        for (int j = 0; j < len; j++) mat[i][j] = 0;
    } else if (type == 1) {
      for (int i = 0; i < len; i++) mat[i][i] = 1;
    }
  }
  matrix operator*(const matrix &b) const {
    matrix c(len, 0);
    for (int i = 0; i < len; i++)
      for (int j = 0; j < len; j++)
        for (int k = 0; k < len; k++)
          c.mat[i][j] =
              (c.mat[i][j] + mat[i][k] * b.mat[k][j] % (mod - 1)) % (mod - 1);
    return c;
  }
};
matrix quick_pow(matrix a, int b) {
  matrix ans = matrix(a.len, 1);
  while (b) {
    if (b & 1) ans = ans * a;
    a = a * a;
    b >>= 1;
  }
  return ans;
}
long long gcd(long long n, long long m) {
  if (m == 0) return n;
  return gcd(m, n % m);
}
long long exgcd(long long n, long long m, long long &x, long long &y) {
  if (m == 0) {
    x = 1;
    y = 0;
    return n;
  }
  long long d = exgcd(m, n % m, y, x);
  y -= n / m * x;
  return d;
}
long long solve(long long a, long long b, long long p) {
  if (b == 0) return 0;
  long long d = gcd(a, p);
  if (b % d) return -1;
  a /= d;
  b /= d;
  p /= d;
  long long x, y;
  exgcd(a, p, x, y);
  x = (x * b) % p;
  if (x < 0) x += p;
  return x;
}
map<long long, long long> M;
namespace BSGS {
long long quick_pow(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans % mod;
}
long long inv(long long x, long long p) { return quick_pow(x, p - 2, p); }
map<long long, long long> tab;
long long bsgs(long long a, long long b, long long p) {
  long long u = (long long)sqrt(p) + 1;
  long long now = 1, step;
  for (long long i = 0; i < u; i++) {
    long long tmp = b * inv(now, p) % p;
    if (!tab.count(tmp)) {
      tab[tmp] = i;
    }
    now = (now * a) % p;
  }
  step = now;
  now = 1;
  for (long long i = 0; i < p; i += u) {
    if (tab.count(now)) {
      return i + tab[now];
    }
    now = (now * step) % p;
  }
  throw;
  return -1;
}
}  // namespace BSGS
const int g = 3;
int main() {
  int n, m, k, x;
  scanf("%d", &k);
  matrix c(k, 0);
  for (int i = 0; i < k; i++) {
    scanf("%d", &x);
    c.mat[0][i] = x % (mod - 1);
  }
  for (int i = 1; i < k; i++) c.mat[i][i - 1] = 1;
  scanf("%d%d", &n, &m);
  c = quick_pow(c, n - k);
  int ans = solve(c.mat[0][0], BSGS::bsgs(g, m, mod), mod - 1);
  if (ans >= 0) {
    printf("%lld\n", BSGS::quick_pow(g, ans, mod));
  } else
    printf("-1\n");
  return 0;
}