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 long long MOD = 998244353;
const long long EULER = 402653184;
const int N = 40000;
long long k, m, n, b[110];
struct mat {
  long long dat[110][110];
  mat() { memset(dat, 0, sizeof dat); }
  mat operator*(mat y) {
    mat tmp;
    for (int i = (1); i <= (k); ++i)
      for (int j = (1); j <= (k); ++j)
        for (int l = (1); l <= (k); ++l) {
          tmp.dat[i][j] += dat[i][l] * y.dat[l][j];
          tmp.dat[i][j] %= (MOD - 1);
        }
    return tmp;
  }
} a;
mat qpow(mat a, int n) {
  mat b;
  for (int i = (1); i <= (100); ++i) b.dat[i][i] = 1;
  for (; n; n >>= 1, a = a * a) {
    if (n & 1) b = b * a;
  }
  return b;
}
long long pw1(long long x, long long n, long long mod) {
  if (n == 0) return 1;
  long long y = pw1(x, n / 2, mod);
  y = y * y % mod;
  if (n & 1)
    return x * y % mod;
  else
    return y;
}
long long GCD(long long x, long long y) {
  long long r = 0;
  while (x) {
    r = y % x, y = x, x = r;
  }
  return y;
}
long long Indr(long long x) {
  long long tmp = 1;
  int M = (int)sqrt(MOD);
  map<long long, int> pws;
  for (int i = (0); i <= (M); ++i) pws[tmp] = i, tmp = tmp * 3 % MOD;
  for (int i = (0); i <= (MOD / M + 1); ++i) {
    tmp = pw1(3, MOD - 1 - i * M % (MOD - 1), MOD) * x % MOD;
    if (pws[tmp]) return pws[tmp] + i * M;
  }
}
int main() {
  cin >> k;
  for (int i = (1); i <= (k); ++i) cin >> b[i];
  for (int i = (1); i <= (k - 1); ++i) a.dat[i][i + 1] = 1;
  for (int i = (1); i <= (k); ++i) a.dat[k][i] = b[k + 1 - i];
  cin >> n >> m;
  a = qpow(a, n - k);
  long long npow = a.dat[k][k];
  long long gcd = GCD(npow, MOD - 1);
  long long invL = pw1(npow / gcd, EULER - 1, MOD - 1);
  long long ind = Indr(m);
  if (ind % gcd != 0) {
    cout << -1 << endl;
    return 0;
  }
  ind /= gcd;
  long long res = pw1(3, ind, MOD);
  cout << pw1(res, invL, MOD) << endl;
  return 0;
}