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;
inline long long read() {
  long long x = 0, f = 1;
  char s = getchar();
  while ((s < '0' || s > '9') && s > 0) {
    if (s == '-') f *= -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 1) + (x << 3) + s - '0';
    s = getchar();
  }
  return x * f;
}
struct matrix {
  int n, m;
  long long c[105][105];
  matrix() { memset(c, 0, sizeof(c)); }
  matrix(int N, int M) {
    n = N, m = M;
    memset(c, 0, sizeof(c));
  }
  matrix operator*(const matrix& a) {
    matrix res = matrix(n, a.m);
    for (int i = 1; i <= n; i++)
      for (int k = 1; k <= m; k++)
        for (int j = 1; j <= a.m; j++)
          res.c[i][j] = (res.c[i][j] + c[i][k] * a.c[k][j]) % (998244353ll - 1);
    return res;
  }
};
inline matrix jzksm(matrix a, long long b) {
  matrix res = matrix(a.n, a.n);
  for (int i = 1; i <= a.n; i++) res.c[i][i] = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    a = a * a;
  }
  return res;
}
inline long long ksm(long long a, long long b, long long mo) {
  long long res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mo;
    a = a * a % mo;
  }
  return res;
}
int k;
long long B[105], n, m;
vector<long long> z;
inline long long BSGS(long long m, long long p) {
  long long sp = (sqrt(p) + 0.5), a = ksm(3, sp, p), b = m;
  map<long long, long long> g;
  for (int i = 1; i <= sp; i++) g[b] = i, b = b * 3 % p;
  b = a;
  for (int i = 1; i <= sp; i++) {
    if (g[b] > 0) return (sp * i % (p - 1) - g[b] + p) % (p - 1);
    b = b * a % p;
  }
  return -1;
}
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;
  }
  long long res = exgcd(b, a % b, y, x);
  y -= a / b * x;
  return res;
}
int main() {
  k = read();
  for (int i = k; i > 0; i--) B[i] = read();
  n = read(), m = read();
  matrix a = matrix(1, k), b = matrix(k, k);
  a.c[1][k] = 1;
  for (int i = 1; i < k; i++) b.c[i + 1][i] = 1;
  for (int i = 1; i <= k; i++) b.c[i][k] = B[i];
  a = a * jzksm(b, n - k);
  if (a.c[1][k] == 0) a.c[1][k] += 998244353ll - 1;
  if (m != 1 && a.c[1][k] == 998244353ll - 1) return 0 * puts("-1");
  if (m == 1) return 0 * puts("1");
  long long t = BSGS(m, 998244353ll), j, o, ans, d;
  d = exgcd(a.c[1][k], 998244353ll - 1, j, o),
  j = (j * (t / d) % (998244353ll - 1) + 998244353ll - 1) % (998244353ll - 1);
  ans = ksm(3, j, 998244353ll);
  if (ksm(ans, a.c[1][k], 998244353ll) == m)
    printf("%lld\n", ans);
  else
    puts("-1");
  return 0;
}