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, mp = mod - 1;
int k, n, m, g[105][105], f[105][105], tmp[105][105];
void mul(int ord) {
  if (ord) {
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= k; j++) tmp[i][j] = f[i][j], f[i][j] = 0;
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= k; j++)
        for (int o = 1; o <= k; o++)
          f[i][j] = (f[i][j] % mp + 1ll * tmp[i][o] * g[o][j] % mp) % mp;
  }
  if (!ord) {
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= k; j++) tmp[i][j] = g[i][j], g[i][j] = 0;
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= k; j++)
        for (int o = 1; o <= k; o++)
          g[i][j] = (g[i][j] % mp + 1ll * tmp[i][o] * tmp[o][j] % mp) % mp;
  }
}
int quickpow(int a, int b) {
  int s = 1;
  while (b) {
    if (b % 2) s = 1ll * s * a % mod;
    a = 1ll * a * a % mod, b /= 2;
  }
  return s;
}
int BSGS(int a, int b, const int p) {
  int m = int(sqrt(p) + 1);
  static map<int, int> has;
  has.clear();
  for (int i = 0, t = b; i < m; i++) has[t] = i, t = 1ll * t * a % p;
  int base = quickpow(a, m), now = 1;
  for (int i = 1; i <= m + 1; i++)
    if (has.count(now = 1ll * now * base % p)) return i * m - has[now];
  return -1;
}
int main() {
  scanf("%d", &k);
  for (int i = 1; i <= k; i++) scanf("%d", &g[i][1]), g[i][i + 1] = 1;
  g[k][k + 1] = 0;
  f[1][1] = 1;
  scanf("%d%d", &n, &m);
  n -= k;
  while (n) {
    if (n % 2) mul(1);
    mul(0);
    n /= 2;
  }
  int ans = BSGS(quickpow(3, f[1][1]), m, mod);
  printf("%d", ans == -1 ? ans : quickpow(3, ans));
}