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;
int proot = 3;
map<int, int> mp;
long long x, y;
long long ex_gcd(long long a, long long b) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long d = ex_gcd(b, a % b);
  long long tmp = x;
  x = y;
  y = tmp - a / b * y;
  return d;
}
long long inv(long long a, long long b) {
  ex_gcd(a, b);
  x %= b;
  if (x <= 0) x += b;
  return x;
}
long long babystep_giantstep(long long a, long long b, long long n) {
  long long m = ceil(sqrt(n * 1.0));
  long long x = 1;
  mp.clear();
  for (int i = 0; i < m; i++) {
    if (!mp.count(x)) mp[x] = i;
    x = x * a % n;
  }
  long long y = inv(x, n);
  x = 1;
  for (int i = 0; i < m; i++) {
    long long tmp = x * b % n;
    if (mp.count(tmp)) return i * m + mp[tmp];
    x = x * y % n;
  }
  return -1;
}
struct matrix {
  int a[110][110];
  matrix() { memset(a, 0, sizeof(a)); }
  matrix operator*(const matrix &p) {
    matrix c;
    for (int i = 0; i < 101; i++)
      for (int j = 0; j < 101; j++)
        for (int k = 0; k < 101; k++)
          if (a[i][k] && p.a[k][j])
            c.a[i][j] = (c.a[i][j] + 1LL * a[i][k] * p.a[k][j]) % (mod - 1);
    return c;
  }
};
matrix mpow(matrix a, int n) {
  matrix b;
  for (int i = 0; i < 101; i++) b.a[i][i] = 1;
  while (n) {
    if (n & 1) b = b * a;
    a = a * a;
    n >>= 1;
  }
  return b;
}
int a[110];
long long fpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
int main() {
  int r;
  scanf("%d", &r);
  for (int i = 0; i < r; i++) scanf("%d", &a[i]);
  int n, m;
  scanf("%d%d", &n, &m);
  if (m == 1) {
    printf("1\n");
    return 0;
  }
  int mm = babystep_giantstep(proot, m, mod);
  matrix c;
  for (int i = 0; i < r; i++) c.a[0][i] = a[i];
  for (int i = 1; i < r; i++) c.a[i][i - 1] = 1;
  c = mpow(c, n - r);
  long long p = c.a[0][0];
  if (!p) {
    printf("-1\n");
    return 0;
  }
  long long d = ex_gcd(p, mod - 1);
  if (mm % d) {
    printf("-1\n");
    return 0;
  }
  p = x * (mm / d) % (mod - 1);
  if (p < 0) p += (mod - 1);
  cout << fpow(3, p) << endl;
  return 0;
}