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 p = 998244353;
const int g = 3;
const int Inf = 2e9;
int GCD(int x, int y) { return y ? GCD(y, x % y) : x; }
void add_self(int& x, int y, const int& m) {
  if ((x += y) >= m) x -= m;
}
void sub_self(int& x, int y, const int& m) {
  if ((x -= y) < 0) x += m;
}
int add(int x, int y, const int& m) { return add_self(x, y, m), x; }
int sub(int x, int y, const int& m) { return sub_self(x, y, m), x; }
int mul(int x, int y, const int& m) { return (long long)x * y % m; }
int fp(int x, int y, const int& m) {
  if (!y) return 1;
  if (y & 1) return mul(x, fp(x, y - 1, m), m);
  return fp(mul(x, x, m), y >> 1, m);
}
int inv(int x, const int& m) { return fp(x, m - 2, m); }
int SolveDiscreteLogarithm(int a, int b, const int& m) {
  assert(0 <= a && a < m && 0 <= b && b < m);
  assert(GCD(a, m) == 1);
  if (!a || !b) return -1;
  if (b == 1) return 0;
  int n = 1;
  while (n * n < m) n++;
  unordered_map<int, int> Max;
  for (int q = 0, x = b; q < n; q++, x = mul(x, a, m)) Max[x] = q;
  int Ans = Inf;
  int an = 1;
  for (int i = 0; i < n; i++) an = mul(an, a, m);
  for (int p = 1, x = an; p * n - (n - 1) <= m; p++, x = mul(x, an, m))
    if (Max.count(x)) Ans = min(Ans, n * p - Max[x]);
  if (Ans == Inf) Ans = -1;
  return Ans;
}
vector<vector<int> > Zero(int n, int m) {
  return vector<vector<int> >(n, vector<int>(m, 0));
}
vector<vector<int> > Ident(int n) {
  vector<vector<int> > Res = Zero(n, n);
  for (int i = 0; i < n; i++) Res[i][i] = 1;
  return Res;
}
vector<vector<int> > Multy(const vector<vector<int> >& A,
                           const vector<vector<int> >& B, const int& m) {
  assert(A[0].size() == B.size());
  vector<vector<int> > Res = Zero(A.size(), B[0].size());
  for (int i = 0; i < Res.size(); i++)
    for (int j = 0; j < Res[0].size(); j++)
      for (int k = 0; k < A[0].size(); k++)
        add_self(Res[i][j], mul(A[i][k], B[k][j], m), m);
  return Res;
}
vector<vector<int> > Power(const vector<vector<int> >& A, int k, const int& m) {
  if (!k) return Ident(A.size());
  if (k & 1) return Multy(A, Power(A, k - 1, m), m);
  return Power(Multy(A, A, m), k >> 1, m);
}
int main() {
  int k;
  cin >> k;
  vector<vector<int> > Base = Zero(1, k);
  vector<vector<int> > Trans = Zero(k, k);
  Base[0][k - 1] = 1;
  for (int i = 0; i < k - 1; i++) Trans[i + 1][i] = 1;
  for (int i = 0; i < k; i++) scanf("%d", &Trans[k - 1 - i][k - 1]);
  int n, m;
  cin >> n >> m;
  vector<vector<int> > Res = Multy(Base, Power(Trans, n - 1, p - 1), p - 1);
  int alpha = Res[0][0];
  int y = SolveDiscreteLogarithm(fp(g, alpha, p), m, p);
  if (y == -1) return 0 * puts("-1");
  int x = fp(g, y, p);
  cout << x << endl;
}