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;
namespace CF1106F {
long long k, n, m;
long long mod = 998244353;
long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') {
      w = -1;
    }
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = (s << 1) + (s << 3) + ch - '0';
    ch = getchar();
  }
  return s * w;
}
struct node {
  long long mat[201][201];
  long long x, y;
} _, s, t;
node operator*(node x, node y) {
  node z;
  z.x = x.x, z.y = y.y;
  for (long long i = 1; i <= x.x; i++) {
    for (long long j = 1; j <= y.y; j++) {
      z.mat[i][j] = 0;
      for (long long k = 1; k <= x.y; k++) {
        z.mat[i][j] = (z.mat[i][j] + x.mat[i][k] * y.mat[k][j]) % (mod - 1);
      }
    }
  }
  return z;
}
node matpow(node x, long long y) {
  node Ans = x;
  y--;
  while (y) {
    if (y & 1) Ans = Ans * x;
    x = x * x;
    y >>= 1;
  }
  return Ans;
}
long long power(long long R, long long Q) {
  long long Ans = 1;
  while (Q) {
    if (Q & 1) Ans = Ans * R % mod;
    R = R * R % mod;
    Q >>= 1;
  }
  return Ans;
}
long long Exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  long long g = Exgcd(b, a % b, x, y);
  long long u = x;
  x = y;
  y = u - (a / b) * x;
  return g;
}
long long BSGS(long long a) {
  m %= mod;
  map<long long, long long> mp;
  long long c = ceil(sqrt(mod)), num = 1, r = 1;
  num = m % mod;
  mp[num] = 0;
  for (long long i = 1; i < c; i++) {
    num = num * a % mod;
    mp[num] = i;
  }
  a = power(a, c);
  if (!a) {
    if (!m)
      return 1;
    else
      return -1;
  }
  if (mp.find(r) != mp.end()) return 0;
  for (long long i = 1; i <= c; i++) {
    r = r * a % mod;
    if (mp.find(r) != mp.end()) return c * i - mp[r];
  }
  return -1;
}
void main() {
  k = read();
  s.x = 1;
  _.x = _.y = s.y = k;
  s.mat[1][k] = 1;
  for (long long i = 2; i <= k; i++) {
    _.mat[i][i - 1] = 1;
  }
  for (long long i = k; i >= 1; i--) {
    _.mat[i][k] = read();
  }
  n = read(), m = read();
  t = s * matpow(_, n - k);
  long long f = BSGS(3);
  long long x, y;
  long long u = Exgcd(t.mat[1][k], mod - 1, x, y);
  if (f % u)
    puts("-1");
  else {
    x = (f / u * x % (mod - 1) + (mod - 1)) % (mod - 1);
    cout << power(3, x) << endl;
  }
  return;
}
}  // namespace CF1106F
int main() {
  CF1106F::main();
  return 0;
}