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;
int N, n, fn;
struct Matrix {
  long long A[105][105];
  Matrix(long long x = 0) {
    memset(A, 0, sizeof(A));
    for (int i = 0; i < 105; ++i) A[i][i] = x;
  }
  const long long* operator[](const int& i) const { return A[i]; }
  long long* operator[](const int& i) { return A[i]; }
  Matrix operator*(const Matrix& M) const {
    Matrix Ans;
    for (int i = 0; i < N; ++i)
      for (int j = 0; j < N; ++j)
        for (int k = 0; k < N; ++k)
          Ans[i][j] = (Ans[i][j] + A[i][k] * M[k][j]) % 998244352;
    return Ans;
  }
  void print() {
    for (int i = 0; i < N; ++i) {
      for (int j = 0; j < N; ++j) cout << A[i][j] << " ";
      cout << endl;
    }
    cout << endl;
  }
} Ans(1), A;
long long exgcd(long long a, long long b, long long& x, long long& y) {
  if (!b) return x = 1, y = 0, a;
  long long d = exgcd(b, a % b, y, x);
  return y -= a / b * x, d;
}
long long pw(long long a, long long k) {
  long long ans = 1;
  for (; k; k >>= 1) {
    if (k & 1) ans = ans * a % 998244353;
    a = a * a % 998244353;
  }
  return ans;
}
long long gk[35005] = {1};
map<long long, long long> Ig;
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N - 1; ++i) A[i][i + 1] = 1;
  for (int i = N - 1; ~i; --i) scanf("%lld", &A[N - 1][i]);
  scanf("%d%d", &n, &fn);
  for (--n; n; n >>= 1) {
    if (n & 1) Ans = Ans * A;
    A = A * A;
  }
  for (int i = 1; i < 35005; ++i) gk[i] = gk[i - 1] * 3 % 998244353;
  for (int i = 0; i < 35005; ++i)
    Ig[gk[i]] = i, assert(pw(3, Ig[gk[i]]) == gk[i]);
  long long b = gk[35005 - 1], Im;
  for (long long t = 1, i = 0; i < 35005; ++i, t = t * b % 998244353) {
    long long q = pw(t, 998244353 - 2) * fn % 998244353;
    if (!Ig.count(q)) continue;
    Im = i * (35005 - 1) + Ig[q];
    break;
  }
  assert(pw(3, Im) == fn);
  long long x, y, k = Ans[0][N - 1], d = exgcd(k, 998244352, x, y);
  if (Im % d)
    puts("-1");
  else {
    x = (x % 998244352 + 998244352) % 998244352;
    long long t = x * (Im / d) % (998244352 / d);
    printf("%lld\n", pw(3, t));
  }
  return 0;
}