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 MOD = 1e9 + 7;
const int MOD998 = 998244353;
const int INF = 1e9;
const long long LLINF = 1e18;
mt19937_64 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
template <class T>
T rnd(T l, T r) {
  return uniform_int_distribution<T>(l, r)(rng);
}
const int K = 105;
using Matrix = array<array<int, K>, K>;
int add(int x, int y, int mod = MOD998) {
  assert(0 <= x && x < mod && 0 <= y && y < mod);
  return x + y >= mod ? x + y - mod : x + y;
}
int sub(int x, int y, int mod = MOD998) {
  assert(0 <= x && x < mod && 0 <= y && y < mod);
  return x - y < 0 ? x - y + mod : x - y;
}
int mul(int x, int y, int mod = MOD998) {
  assert(0 <= x && x < mod && 0 <= y && y < mod);
  return 1LL * x * y % mod;
}
int pw(int a, int n, int mod = MOD998) {
  assert(0 <= a && a < mod);
  int res = 1 % mod;
  while (n) {
    if (n & 1) {
      res = mul(res, a, mod);
    }
    a = mul(a, a, mod);
    n >>= 1;
  }
  return res;
}
int inv(int x, int mod = MOD998) { return pw(x, mod - 2, mod); }
vector<int> get_invs(int upto, int mod = MOD998) {
  vector<int> res(upto + 1);
  res[1] = 1;
  for (int i = 2; i <= upto; i++) {
    res[i] = mod - mul(mod / i, res[mod % i], mod);
  }
  return res;
}
Matrix add(Matrix a, Matrix b) {
  Matrix c;
  for (int i = 0; i < K; i++) {
    for (int j = 0; j < K; j++) {
      c[i][j] = add(a[i][j], b[i][j], MOD998 - 1);
    }
  }
  return c;
}
Matrix empty() {
  Matrix res;
  for (int i = 0; i < K; i++) {
    res[i].fill(0);
  }
  return res;
}
Matrix mul(Matrix a, Matrix b) {
  Matrix c = empty();
  for (int i = 0; i < K; i++) {
    for (int j = 0; j < K; j++) {
      for (int k = 0; k < K; k++) {
        c[i][j] = add(c[i][j], mul(a[i][k], b[k][j], MOD998 - 1), MOD998 - 1);
      }
    }
  }
  return c;
}
Matrix eye() {
  Matrix res = empty();
  for (int i = 0; i < K; i++) {
    res[i][i] = 1;
  }
  return res;
}
Matrix pw(Matrix a, int n) {
  Matrix res = eye();
  for (; n; n >>= 1, a = mul(a, a))
    if (n & 1) {
      res = mul(res, a);
    }
  return res;
}
void run() {
  int k;
  cin >> k;
  vector<int> b(k);
  for (int &x : b) {
    cin >> x;
    x %= (MOD998 - 1);
  }
  int n, m;
  cin >> n >> m;
  Matrix A = empty();
  for (int i = 0; i < k - 1; i++) {
    A[i][i + 1] = 1;
  }
  for (int i = 0; i < k; i++) {
    A[k - 1][i] = b[k - i - 1];
  }
  A = pw(A, n - k);
  int power = A[k - 1][k - 1];
  int root = 3;
  int base = pw(root, power);
  if (!base) {
    cout << "-1\n";
    return;
  }
  int N = 31595;
  int P = (MOD998 + N - 1) / N;
  map<int, int> fp;
  for (int i = 0; i < P; i++) {
    int deg = N * (i + 1) % (MOD998 - 1);
    fp[pw(base, deg)] = N * (i + 1);
  }
  for (int i = 0; i <= N; i++) {
    int fq = mul(m, pw(base, i));
    if (fp.count(fq)) {
      cout << pw(root, (fp[fq] - i)) << "\n";
      return;
    }
  }
  cout << "-1\n";
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  int tests = 1;
  while (tests--) {
    run();
  }
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s.\n";
  return 0;
}