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;
long long int modulo = 998244353;
long long int phi = modulo - 1;
int n;
vector<vector<long long int> > operator*(vector<vector<long long int> > m1,
                                         vector<vector<long long int> > m2) {
  vector<vector<long long int> > m(n, vector<long long int>(n, 0));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      for (int k = 0; k < n; k++)
        m[i][j] = (m[i][j] + m1[i][k] * m2[k][j]) % phi;
  return m;
}
vector<long long int> operator*(vector<vector<long long int> > m,
                                vector<long long int> v) {
  vector<long long int> r(n, 0);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) r[i] = (r[i] + m[i][j] * v[j]) % phi;
  return r;
}
vector<vector<long long int> > expo(vector<vector<long long int> > m,
                                    long long int e) {
  if (e == 0) {
    vector<vector<long long int> > id(n, vector<long long int>(n, 0));
    for (int i = 0; i < n; i++) id[i][i] = 1;
    return id;
  }
  if (e % 2 == 0) return expo(m * m, e / 2);
  return m * expo(m * m, e / 2);
}
long long int absolut(long long int x) {
  if (x < 0) return -x;
  return x;
}
long long int gcd(long long int a, long long int b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  return gcd(a % b, b);
}
long long int mcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
void chino(long long int a, long long int b, long long int &ca,
           long long int &cb, long long int d) {
  if (absolut(a) < absolut(b)) {
    chino(b, a, cb, ca, d);
    return;
  }
  if (b == 0) {
    if (d % a != 0) {
      cout << -1 << endl;
      exit(0);
    }
    cb = 0;
    ca = d / a;
    return;
  }
  chino(a % b, b, ca, cb, d);
  cb -= ca * (a / b);
}
long long int expo(long long int b, long long int e) {
  if (e == 0) return 1;
  if (e % 2 == 0) return expo(b * b % modulo, e / 2);
  return b * expo(b * b % modulo, e / 2) % modulo;
}
long long int inverse(long long int x) { return expo(x, modulo - 2); }
long long int getexpo3(long long int b) {
  long long int sq = 1;
  while (sq * sq < modulo) sq++;
  long long int inv3 = inverse(3);
  map<long long int, int> mapa;
  {
    mapa[b] = 0;
    long long int val = b;
    for (int i = 1; i < sq; i++) {
      val = val * inv3 % modulo;
      mapa[val] = i;
    }
  }
  {
    long long int e3tosq = expo(3, sq);
    long long int val = 1;
    for (int i = 0; i < sq; i++) {
      if (mapa.count(val)) return (i * sq + mapa[val]) % phi;
      val = e3tosq * val % modulo;
    }
  }
  return 0;
}
int N, FN, K;
int main() {
  cin >> K;
  n = K;
  vector<vector<long long int> > m(n, vector<long long int>(n, 0));
  vector<long long int> v(n, 0);
  v[0] = 1;
  for (int i = 0; i < n; i++) cin >> m[0][i];
  for (int i = 1; i < n; i++) m[i][i - 1] = 1;
  cin >> N >> FN;
  long long int E = (expo(m, N - K) * v)[0];
  long long int e3 = getexpo3(FN);
  long long int cE, cphi;
  chino(E, phi, cE, cphi, e3);
  if (E == 0) {
    cout << 1 << endl;
    exit(0);
  }
  long long int desp = mcm(E, phi) / E;
  cE %= desp;
  if (cE < 0) cE += desp;
  long long int sol = expo(3, cE);
  cout << sol << endl;
}