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 mod = 998244353;
const int g = 3;
long long qpow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, y, x);
  y -= (a / b) * x;
  return d;
}
long long inv(long long a, long long m) {
  long long x1, x2;
  long long d = exgcd(a, m, x1, x2);
  if (d > 1) return -1;
  x1 = ((x1 % m) + m) % m;
  return x1;
}
long long ex_bsgs(long long a, long long b, long long m) {
  if (a % m == 0 && b) return -1;
  long long k = 0;
  long long ak = 1;
  long long d;
  while ((d = gcd(a, m)) > 1) {
    if (b % d) return -1;
    m /= d, b /= d;
    ak = ak * (a / d) % m;
    k++;
    if (ak == b) return k;
  }
  long long inva = inv(ak, m);
  b = inva * b % m;
  long long n = ceil(sqrt(m));
  long long bn = b;
  map<long long, int> MP;
  for (int i = 0; i <= n; i++) {
    MP[bn] = i;
    bn = bn * a % m;
  }
  long long an = qpow(a, n, m);
  long long atn = 1;
  for (int i = 1; i <= n; i++) {
    atn = atn * an % m;
    if (MP.find(atn) != MP.end()) return i * n - MP[atn] + k;
  }
  return -1;
}
const int maxn = 105;
struct matrix {
  int r, c;
  int m[maxn][maxn];
  matrix(int a, int b) {
    r = a, c = b;
    memset(m, 0, sizeof(m));
  }
  matrix(const matrix &m1) {
    r = m1.r, c = m1.c;
    for (int i = 1; i <= r; i++)
      for (int j = 1; j <= c; j++) m[i][j] = m1.m[i][j];
  }
  matrix operator*(const matrix &b) const {
    matrix ans(r, b.c);
    for (int i = 1; i <= r; i++) {
      for (int j = 1; j <= b.c; j++) {
        for (int k = 1; k <= c; k++) {
          ans.m[i][j] += (1ll * m[i][k] * b.m[k][j]) % (mod - 1);
          if (ans.m[i][j] >= (mod - 1)) ans.m[i][j] -= (mod - 1);
        }
      }
    }
    return ans;
  }
  matrix qpow(int q) {
    matrix ans(r, c);
    matrix a = *this;
    for (int i = 1; i <= a.r; i++) ans.m[i][i] = 1;
    while (q) {
      if (q & 1) ans = ans * a;
      q >>= 1;
      a = a * a;
    }
    return ans;
  }
};
int main() {
  int k;
  cin >> k;
  vector<int> b(k + 1);
  for (int i = 1; i <= k; i++) cin >> b[i];
  long long n, m;
  cin >> n >> m;
  long long hn = ex_bsgs(g, m, mod);
  matrix M(k, k);
  for (int i = 2; i <= k; i++) M.m[i][i - 1] = 1;
  for (int i = 1; i <= k; i++) M.m[1][i] = b[i];
  matrix ans = M.qpow((n - k));
  long long x1, x2;
  long long d = exgcd(ans.m[1][1], mod - 1, x1, x2);
  if (hn % d) {
    cout << -1 << endl;
    return 0;
  }
  x1 *= (hn) / d;
  long long wd = (mod - 1) / d;
  x1 = (x1 % wd + wd) % wd;
  long long res = qpow(g, x1, mod);
  cout << res << endl;
  return 0;
}