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;
mt19937 rnd(time(nullptr));
const long long MAXN = 110, M = 998244353, SQ = 30000;
long long n, m, k, N;
struct Matrix {
  int a[MAXN][MAXN];
  Matrix& operator*(const Matrix& b) const {
    Matrix& rez = *(new Matrix);
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        rez.a[i][j] = 0;
        for (int z = 0; z < N; z++) {
          rez.a[i][j] = (rez.a[i][j] + 1LL * a[i][z] * b.a[z][j]) % (M - 1);
        }
      }
    }
    return rez;
  }
};
Matrix& st(Matrix& a, int b) {
  if (b == 1) return a;
  if (b == 2) return a * a;
  if (b % 2) return a * st(a * a, b / 2);
  return st(a * a, b / 2);
}
long long st(long long a, int b) {
  if (b == 0) return 1;
  if (b % 2) return (a * st((a * a) % M, b / 2)) % M;
  return (st((a * a) % M, b / 2)) % M;
}
long long tta = 1, ttb = 0, ta = 0, tb = 1;
int Mygcd(int a, int b) {
  while (b != 0) {
    int q = a / b;
    int r = a - q * b;
    tta -= ta * q;
    ttb -= tb * q;
    swap(ta, tta);
    swap(tb, ttb);
    ta %= M - 1;
    tb %= M - 1;
    tta %= M - 1;
    ttb %= M - 1;
    a = b;
    b = r;
  }
  return a;
}
int ind(int a, int b) {
  long long n1 = M / SQ + 10, n2 = SQ, tc;
  set<pair<long long, int>>::iterator it;
  set<pair<long long, int>> f;
  for (int i = 0; i < n1; ++i) {
    f.insert({st(a, (SQ * i) % (M - 1)), i});
  }
  for (int j = 0; j < n2; j++) {
    tc = (b * st(a, j)) % M;
    it = f.lower_bound({tc, -1});
    if (it != f.end() && (*it).first == tc) {
      int ans = (*it).second * SQ - j;
      ans %= M - 1;
      ans += M - 1;
      ans %= M - 1;
      return ans;
    }
  }
  throw;
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> k;
  N = k;
  Matrix ma;
  for (int i = 0; i < k; i++) {
    cin >> ma.a[k - i - 1][k - 1];
  }
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < k - 1; j++) {
      if (i == j + 1)
        ma.a[i][j] = 1;
      else
        ma.a[i][j] = 0;
    }
  }
  cin >> n >> m;
  long long in = st(ma, n - k).a[N - 1][N - 1];
  long long in2 = ind(3, m);
  long long gc = Mygcd(in, M - 1);
  if (in2 % gc != 0) {
    cout << -1;
    return 0;
  } else {
    tta = (tta % (M - 1) + M - 1) % (M - 1);
    tta = (tta * (in2 / gc)) % (M - 1);
    cout << st(3, tta);
  }
  return 0;
}