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 ll = long long;
ll modpow(ll x, ll n, ll mod) {
  ll res = 1;
  while (n > 0) {
    if (n & 1) (res *= x) %= mod;
    (x *= x) %= mod;
    n >>= 1;
  }
  return res;
}
ll baby_step_giant_step(ll a, ll b, ll p) {
  ll res = -1;
  const ll m = std::sqrt(p) + 1;
  std::map<ll, ll> table;
  for (ll i = 0, z = 1, t = modpow(a, m, p); i < m; ++i) {
    table[z] = i;
    z = (z * t) % p;
  }
  ll z = b * modpow(a, p - m, p) % p;
  for (int i = m - 1; i >= 0; --i) {
    if (table.count(z) == 1) {
      res = m * table[z] + i;
      break;
    }
    z = (z * a) % p;
  }
  return res;
}
using namespace std;
using matrix = vector<vector<ll>>;
constexpr int mod = 998244353;
matrix operator*(matrix const& a, matrix const& b) {
  const int n = a.size();
  matrix res(n, vector<ll>(n));
  for (int k = 0; k < n; ++k) {
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        (res[i][j] += a[i][k] * b[k][j]) %= (mod - 1);
      }
    }
  }
  return res;
}
matrix matpow(matrix x, ll n) {
  matrix res(x.size(), vector<ll>(x.size()));
  for (int i = 0; i < (int)x.size(); ++i) {
    res[i][i] = 1;
  }
  while (n > 0) {
    if (n & 1) res = res * x;
    x = x * x;
    n >>= 1;
  }
  return res;
}
int main() {
  int k;
  cin >> k;
  vector<ll> b(k);
  for (int i = 0; i < k; ++i) {
    cin >> b[i];
  }
  ll n, m;
  cin >> n >> m;
  matrix A;
  A.push_back(b);
  for (int i = 0; i < k - 1; ++i) {
    vector<ll> a(k);
    a[i] = 1;
    A.push_back(a);
  }
  const ll s = matpow(A, n - k)[0][0];
  const ll x = modpow(3, s, mod);
  const ll z = baby_step_giant_step(x, m, mod);
  if (z < 0) {
    cout << -1 << endl;
  } else {
    cout << modpow(3, z, mod) << endl;
  }
}