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 int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - 48;
    s = getchar();
  }
  return f * x;
}
const int p = 998244353, g = 3;
int k, n, m;
struct mat {
  int a[105][105];
  int n, m;
  void clear() {
    memset(a, 0, sizeof(a));
    n = m = 0;
  }
} base, ans;
mat operator*(mat a, mat b) {
  mat ans;
  ans.clear();
  ans.n = a.n, ans.m = b.m;
  for (int i = 1; i <= a.n; i++)
    for (int j = 1; j <= b.m; j++)
      for (int k = 1; k <= a.m; k++)
        ans.a[i][j] =
            (ans.a[i][j] + 1ll * a.a[i][k] * b.a[k][j] % (p - 1)) % (p - 1);
  return ans;
}
inline int ksm(int a, int b, int p) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % p)
    if (b & 1) ans = 1ll * ans * a % p;
  return ans;
}
map<int, int> Hash;
int bsgs(int a, int b, int p) {
  Hash.clear();
  b %= p;
  int m = ceil(sqrt(1.0 * p));
  for (int i = 1; i <= m; i++) {
    b = 1ll * b * a % p;
    Hash[b] = i;
  }
  int tmp = ksm(a, m, p);
  b = 1;
  for (int i = 1; i <= m; i++) {
    b = 1ll * b * tmp % p;
    if (Hash[b]) return ((1ll * i * m - Hash[b]) % p + p) % p;
  }
  return -1;
}
int exgcd(int a, int b, int &x, int &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  int ans = exgcd(b, a % b, y, x);
  y -= a / b * x;
  return ans;
}
int main() {
  k = read();
  ans.clear(), base.clear();
  base.n = base.m = k;
  for (int i = 1; i <= k; i++) base.a[i][1] = read();
  for (int i = 2; i <= k; i++) base.a[i - 1][i] = 1;
  n = read() - k, m = read();
  ans.n = 1, ans.m = k;
  ans.a[1][1] = 1;
  for (; n; n >>= 1, base = base * base)
    if (n & 1) ans = ans * base;
  int t = ans.a[1][1];
  int res = bsgs(g, m, p);
  int x, y;
  int GCD = exgcd((p - 1), t, x, y);
  y = (y % (p - 1) + p - 1) % (p - 1);
  if (res % GCD != 0)
    puts("-1");
  else {
    x = 1ll * y * (res / GCD) % (p - 1);
    int ans = ksm(g, x, p);
    printf("%d\n", ans);
  }
  return 0;
}