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 primo = 998244353;
const long long m = 110;
const long long mod = primo - 1;
long long x;
struct Matrix {
  long long mat[m][m];
  Matrix operator*(const Matrix &p) {
    Matrix ans;
    for (long long i = 0; i < x; i++)
      for (long long j = 0; j < x; j++)
        for (long long k = ans.mat[i][j] = 0; k < x; k++)
          ans.mat[i][j] = (ans.mat[i][j] + mat[i][k] * p.mat[k][j]) % mod;
    return ans;
  }
};
Matrix fExp(Matrix a, long long b) {
  Matrix ans;
  for (long long i = 0; i < m; i++)
    for (long long j = 0; j < m; j++) ans.mat[i][j] = i == j;
  while (b) {
    if (b & 1) ans = ans * a;
    a = a * a;
    b >>= 1;
  }
  return ans;
}
long long fexp(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
long long discreteLog(long long a, long long b, long long m) {
  a %= m;
  b %= m;
  long long n = (long long)sqrt(m + .0) + 1, an = 1;
  for (long long i = 0; i < n; i++) {
    an = (an * a) % m;
  }
  map<long long, long long> vals;
  for (long long i = 1, cur = an; i <= n; i++) {
    if (!vals.count(cur)) vals[cur] = i;
    cur = (cur * an) % m;
  }
  long long ans = 1e18;
  for (long long i = 0, cur = b; i <= n; i++) {
    if (vals.count(cur)) {
      ans = min(ans, vals[cur] * n - i);
    }
    cur = (cur * a) % m;
  }
  return ans;
}
long long primitiveRoot(long long p) {
  vector<long long> fact;
  long long phi = p - 1, n = phi;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      fact.push_back(i);
      while (n % i == 0) {
        n /= i;
      }
    }
  }
  if (n > 1) {
    fact.push_back(n);
  }
  for (long long res = 2; res <= p; res++) {
    bool ok = true;
    for (auto it : fact) {
      ok &= fexp(res, phi / it, p) != 1;
      if (!ok) {
        break;
      }
    }
    if (ok) {
      return res;
    }
  }
  return -1;
}
long long discreteRoot(long long k, long long a, long long mod) {
  long long g = primitiveRoot(mod);
  long long y = discreteLog(fexp(g, k, mod), a, mod);
  if (y == -1) {
    return y;
  }
  return fexp(g, y, mod);
}
int32_t main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  cin >> x;
  Matrix base;
  for (long long i = 0; i < x; i++)
    for (long long j = 0; j < x; j++) {
      base.mat[i][j] = (i == j + 1);
    }
  for (long long i = 0; i < x; i++) cin >> base.mat[0][i];
  long long n, res;
  cin >> n >> res;
  base = fExp(base, n - 1);
  long long exp = base.mat[x - 1][0];
  long long ans = discreteRoot(exp, res, primo);
  cout << (fexp(ans, exp, primo) == res ? ans : -1) << endl;
}