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 INF = (long long)1000000007 * 1000000007;
const long double eps = 1e-11;
const long double pi = acos(-1.0);
vector<vector<long long>> mtmul(vector<vector<long long>> &A,
                                vector<vector<long long>> &B) {
  vector<vector<long long>> C(A.size(), vector<long long>(B[0].size()));
  for (int i = 0; i < (int)A.size(); i++) {
    for (int k = 0; k < (int)B.size(); k++) {
      for (int j = 0; j < (int)B[0].size(); j++) {
        C[i][j] = (C[i][j] + (A[i][k] * B[k][j]) % (mod - 1)) % (mod - 1);
      }
    }
  }
  return C;
}
vector<vector<long long>> mtpow(vector<vector<long long>> A, long long n) {
  vector<vector<long long>> B(A.size(), vector<long long>(A.size()));
  for (int i = 0; i < (int)A.size(); i++) {
    B[i][i] = 1;
  }
  while (n > 0) {
    if (n & (long long)1) B = mtmul(B, A);
    A = mtmul(A, A);
    n >>= 1;
  }
  return B;
}
long long mod_pow(long long a, long long n, long long m) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * a % m;
    n >>= 1;
    a = a * a % m;
  }
  return res;
}
long long baby_step_giant_step_algorithm(long long a, long long b,
                                         long long p) {
  long long res = -1;
  long long rp = sqrt(p) + 1;
  map<long long, bool> exi;
  map<long long, long long> num;
  long long t = mod_pow(a, rp, p);
  long long z = 1;
  for (int i = 0; i < rp; i++) {
    exi[z] = true;
    num[z] = i;
    z = z * t % p;
  }
  z = b * mod_pow(a, p - rp, p) % p;
  for (int i = rp - 1; i >= 0; i--) {
    if (exi[z]) {
      res = rp * num[z] + i;
      break;
    }
    z = z * a % p;
  }
  return res;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int k;
  cin >> k;
  vector<vector<long long>> A;
  vector<long long> b;
  b.resize(k);
  for (int i = 0; i < k; i++) {
    cin >> b[i];
  }
  A.push_back(b);
  for (int i = 1; i <= k - 1; i++) {
    vector<long long> v;
    for (int j = 0; j < k; j++) {
      if (i - j == 1) {
        v.push_back(1);
      } else
        v.push_back(0);
    }
    A.push_back(v);
  }
  long long n, m;
  cin >> n >> m;
  A = mtpow(A, n - k);
  long long sum = A[0][0];
  long long x = mod_pow(3, sum, mod);
  long long y = m;
  long long z = baby_step_giant_step_algorithm(x, y, mod);
  if (z < 0) {
    cout << -1 << endl;
  } else {
    cout << mod_pow(3, z, mod) << endl;
  }
  return 0;
}