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, SQ = 31596, MOD2 = 998244352;
vector<vector<long long> > ones(int n) {
  vector<vector<long long> > r(n, vector<long long>(n));
  for (int i = (0), qwerty = (n); i < qwerty; i++) r[i][i] = 1;
  return r;
}
long long mult(long long a, long long b, long long m = MOD) {
  return (a * b) % m;
}
vector<vector<long long> > operator*(vector<vector<long long> > &a,
                                     vector<vector<long long> > &b) {
  int n = int((a).size()), m = int((b[0]).size()), z = int((a[0]).size());
  vector<vector<long long> > r(n, vector<long long>(m));
  for (int i = (0), qwerty = (n); i < qwerty; i++)
    for (int j = (0), qwerty = (m); j < qwerty; j++)
      for (int k = (0), qwerty = (z); k < qwerty; k++)
        r[i][j] += a[i][k] * b[k][j], r[i][j] %= MOD2;
  return r;
}
vector<vector<long long> > be(vector<vector<long long> > b, long long e) {
  vector<vector<long long> > r = ones(int((b).size()));
  while (e)
    if (e & 1)
      r = r * b, e ^= 1;
    else
      b = b * b, e >>= 1;
  return r;
}
long long be(long long b, long long e, long long m = MOD) {
  long long r = 1;
  while (e)
    if (e & 1)
      r = (r * b) % m, e ^= 1;
    else
      b = (b * b) % m, e >>= 1;
  return r;
}
const long long rt = 3;
int k;
long long n, m;
long long discrete_log(long long b, long long a) {
  map<long long, long long> ds;
  for (int x = (0), qwerty = (SQ); x < qwerty; x++)
    ds[mult(a, be(be(b, MOD - 2), x))] = x;
  for (int x = (0), qwerty = (SQ); x < qwerty; x++) {
    long long ts = be(b, SQ * x);
    if (ds.count(ts)) return (SQ * x + ds[ts]) % MOD2;
  }
  return -1;
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> k;
  vector<vector<long long> > b(k, vector<long long>(k));
  for (int x = (0), qwerty = (k); x < qwerty; x++) cin >> b[0][x];
  for (int x = (1), qwerty = (k); x < qwerty; x++) b[x][x - 1] = 1;
  cin >> n >> m;
  b = be(b, n - k);
  long long ts = discrete_log(3, m);
  while (ts == -1)
    ;
  long long phi_mod1 = 402653184;
  long long g = gcd(ts, b[0][0]);
  if (g) {
    ts /= g;
    b[0][0] /= g;
  } else {
    cout << "1\n";
    exit(0);
  }
  if (ts == -1 || gcd(MOD - 1, b[0][0]) > 1)
    cout << "-1\n";
  else
    cout << be(3, (ts * be(b[0][0], phi_mod1 - 1, MOD - 1)) % (MOD - 1))
         << "\n";
}