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;
int a[205], b[205], mo[205], tmp[205], n, k, m, ans;
int q_pow(int x, int n) {
  int ret = 1;
  for (; n; n >>= 1, x = (long long)x * x % 998244353)
    if (n & 1) ret = (long long)ret * x % 998244353;
  return ret;
}
void mul(int *a, int *b, int *ret) {
  memset(tmp, 0, sizeof(tmp));
  for (int i = 0; i < k; i++)
    for (int j = 0; j < k; j++)
      tmp[i + j] = (tmp[i + j] + (long long)a[i] * b[j]) % 998244352;
  for (int i = (k << 1) - 2; i >= k; i--)
    if (tmp[i])
      for (int j = 1; j <= k; j++)
        tmp[i - j] = (tmp[i - j] - (long long)tmp[i] * mo[k - j]) % 998244352;
  for (int i = 0; i < k; i++) ret[i] = tmp[i];
}
map<int, int> mp;
int ex_gcd(int a, int b, int &x, int &y) {
  if (!b) return x = 1, y = 0, a;
  int ret = ex_gcd(b, a % b, y, x);
  y = y - a / b * x;
  return ret;
}
int BSGS(int x) {
  int s = 10000, t = 1;
  for (int i = 0; i < s; i++)
    mp[int((long long)t * x % 998244353)] = i, t = t * 3ll % 998244353;
  for (int i = 1, now = 1;; i++) {
    now = (long long)now * t % 998244353;
    if (mp.find(now) != mp.end()) return i * s - mp[now];
  }
}
int main() {
  scanf("%d", &k);
  for (int i = 1; i <= k; i++)
    scanf("%d", &mo[k - i]), mo[k - i] = 998244352 - mo[k - i];
  scanf("%d%d", &n, &m);
  b[1] = 1;
  a[0] = 1;
  if (k == 1) b[0] = 998244352 - mo[0], b[1] = 0;
  for (n--; n; n >>= 1) {
    if (n & 1) mul(a, b, a);
    mul(b, b, b);
  }
  int t = (a[k - 1] + 998244352) % 998244352, x = 0, y = 0;
  int tt = ex_gcd(t, 998244352, x, y), kk = BSGS(m);
  if (kk % tt) return puts("-1"), 0;
  x = ((long long)x * (kk / tt) % 998244352 + 998244352) % 998244352;
  printf("%d\n", q_pow(3, x));
}