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;
const int phi = 998244352;
const int g = 3;
struct mat {
  int n, m;
  long long a[105][105];
  mat() {
    n = m = 0;
    memset(a, 0, sizeof(a));
  }
  void pr() {
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) cout << a[i][j] << ' ';
      cout << '\n';
    }
  }
  friend mat operator*(const mat &a, const mat &b) {
    mat c;
    c.n = a.n;
    c.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++)
          c.a[i][j] = (c.a[i][j] + a.a[i][k] * b.a[k][j] % phi) % phi;
    return c;
  }
} acc, bas;
int k;
int n, m;
int b[105];
long long qkpow(int a, long long b) {
  if (b == -1) return -1;
  if (b == 0) return 1;
  if (b == 1) return a;
  long long t = qkpow(a, b / 2);
  t = t * t % mod;
  if (b & 1) t = t * a % mod;
  return t;
}
mat qkpow(mat a, long long b) {
  if (b == 1) return a;
  mat t = qkpow(a, b / 2);
  t = t * t;
  if (b & 1) t = t * a;
  return t;
}
long long bsgs(long long a, long long b, long long mod) {
  unordered_map<int, int> f;
  long long m = ceil(sqrt(mod));
  long long temp = 1;
  for (int i = 1; i <= m; i++) {
    temp = temp * a % mod;
    f[b * temp % mod] = i;
  }
  long long ret = temp;
  for (int i = 1; i <= m; i++, ret = ret * temp % mod)
    if (f.count(ret)) return i * m - f[ret];
  return -1;
}
int main() {
  cin >> k;
  for (int i = 1; i <= k; i++) cin >> b[i];
  cin >> n >> m;
  bas.n = k;
  bas.m = 1;
  bas.a[k][1] = 1;
  acc.n = acc.m = k;
  for (int i = 2; i <= k; i++) acc.a[i - 1][i] = 1;
  for (int i = 1; i <= k; i++) acc.a[k][i] = b[k - i + 1];
  bas = qkpow(acc, n - k) * bas;
  long long now = qkpow(3, bas.a[k][1]);
  cout << qkpow(3, bsgs(now, m, mod));
  return 0;
}