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;
struct Matrix {
  int a[105][105];
  int* operator[](int x) { return a[x]; }
  Matrix() { memset(a, 0, sizeof(a)); }
  Matrix operator*(Matrix b) {
    Matrix c;
    for (int i = 0; i < 100; ++i)
      for (int j = 0; j < 100; ++j)
        for (int k = 0; k < 100; ++k)
          c[i][j] = (c[i][j] + 1ll * a[i][k] * b[k][j]) % (MOD - 1);
    return c;
  }
};
int solve1(int b[], int k, int n, int m) {
  Matrix base, ans;
  for (int i = 1; i < k; ++i) base[i][i - 1] = 1;
  for (int i = 1; i <= k; ++i) base[k - i][k - 1] = b[i];
  n -= k;
  ans[0][k - 1] = 1;
  for (; n; n >>= 1, base = base * base)
    if (n & 1) ans = ans * base;
  return ans[0][k - 1];
}
int bsgs(int a, int b, int c, int m) {
  unordered_map<int, int> h;
  if (a == c) return 0;
  int q = 1 << 15, b2 = 1;
  for (int i = 1; i <= q; ++i) {
    c = 1ll * c * b % MOD;
    b2 = 1ll * b2 * b % MOD;
    h[c] = i;
  }
  for (int i = 1; i <= m / q + 1; ++i) {
    a = 1ll * a * b2 % MOD;
    int k = h[a];
    if (k != 0) return i * q - k;
  }
  return -1;
}
inline int fpow(int x, int n, int ret = 1) {
  for (; n; n >>= 1, x = 1ll * x * x % MOD)
    n& 1 ? ret = 1ll * ret * x % MOD : 0;
  return ret;
}
int b[(int)1e5 + 5];
int main() {
  int n, m, k;
  scanf("%d", &k);
  for (int i = 1; i <= k; ++i) scanf("%d", &b[i]);
  scanf("%d%d", &n, &m);
  int c = solve1(b, k, n, m);
  c = fpow(3, c);
  int d = bsgs(1, c, m, MOD);
  if (d == -1)
    puts("-1");
  else
    printf("%d\n", fpow(3, d));
  return 0;
}