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 D = (int)101, mod = (int)998244353, N = D;
long long fp(long long x, long long k) {
  if (k == 0) return 1;
  long long hf = fp(x, k / 2);
  return k % 2 ? hf * hf % mod * x % mod : hf * hf % mod;
}
struct matrix {
  const static int mod = 998244352;
  int size;
  long long a[D][D];
  void init(int _size = -1) {
    if (_size != -1) size = _size;
    memset(a, 0, sizeof(a));
  }
  matrix() { (*this).init(0); }
  matrix(int _size = -1) { (*this).init(_size); }
  void set_identity() {
    for (int i = 0; i < size; i++) {
      for (int j = 0; j < size; j++) a[i][j] = (long long)(i == j);
    }
  }
  matrix operator*(const matrix &B) const {
    matrix C = matrix(size);
    for (int i = 0; i < size; i++)
      for (int j = 0; j < size; j++) {
        for (int k = 0; k < size; k++) C.a[i][j] += a[i][k] * B.a[k][j] % mod;
        C.a[i][j] %= mod;
      }
    return C;
  }
  matrix operator^(const long long &p) const {
    matrix A = (*this), res = matrix(A.size);
    res.set_identity();
    long long t = p;
    while (t > 0) {
      if (t % 2) res = res * A;
      A = A * A;
      t /= 2;
    }
    return res;
  }
};
struct ExtGCD {
  long long extgcd(long long a, long long b, long long &x, long long &y) {
    long long d = a;
    if (b != 0) {
      d = extgcd(b, a % b, y, x);
      y -= (a / b) * x;
    } else {
      x = 1;
      y = 0;
    }
    return d;
  }
  long long find(long long a, long long b, long long res) {
    long long x, y;
    long long g = extgcd(a, b, x, y);
    if (res % g != 0) return -1;
    return x * (res / g);
  }
};
struct BSGS {
  const static int p = 998244353;
  const static int g = 3;
  int m;
  map<int, int> mp;
  vector<int> V;
  void pre() {
    m = (int)ceil(sqrt(p + 0.0) + 1);
    int cg = 1, revgm = 1;
    int invm = (int)(fp(fp(g, m), p - 2));
    for (int i = 0; i < m; i++) {
      mp[cg] = i;
      V.push_back(revgm);
      cg = (int)(1LL * cg * g % p);
      revgm = (int)(1LL * revgm * invm % p);
    }
  }
  int find(int y) {
    for (int i = 0; i < m; i++) {
      int cur = (int)(1LL * V[i] * y % p);
      if (mp.count(cur)) {
        return i * m + mp[cur];
      }
    }
    return -1;
  }
};
int k;
long long b[N];
int n, x;
int main() {
  cin >> k;
  for (int i = 0; i < k; i++) cin >> b[i];
  cin >> n >> x;
  matrix mt = matrix(k);
  for (int i = 0; i < k; i++) mt.a[0][i] = b[i];
  for (int i = 1; i < k; i++) mt.a[i][i - 1] = 1;
  mt = mt ^ (n - k);
  long long c = mt.a[0][0];
  BSGS bsgs;
  bsgs.pre();
  int hn = bsgs.find(x);
  ExtGCD gcd;
  long long hk = gcd.find(c, mod - 1, hn);
  if (hk == -1) return !printf("-1");
  hk = (hk % (mod - 1) + (mod - 1)) % (mod - 1);
  assert(1LL * c * hk % (mod - 1) == 1LL * hn % (mod - 1));
  long long res = fp(3, hk);
  cout << res << endl;
}