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 N = 1e2 + 10;
const long long Mod = 998244353;
long long add(long long a, long long b, long long mod = Mod) {
  return (a += b) >= mod ? a - mod : a;
}
long long sub(long long a, long long b, long long mod = Mod) {
  return (a -= b) < 0 ? a + mod : a;
}
long long mul(long long a, long long b, long long mod = Mod) {
  return 1ll * a * b % mod;
}
long long fast_pow(long long a, long long b, long long mod = Mod) {
  long long res = 1;
  for (; b; b >>= 1, a = mul(a, a, mod))
    if (b & 1) res = mul(res, a, mod);
  return res;
}
long long n, m, k, b[N];
struct Matrix {
  long long g[N][N];
  Matrix() { memset(g, 0, sizeof(g)); }
};
Matrix operator*(const Matrix a, const Matrix b) {
  Matrix c;
  for (long long i = 1; i <= k; i++)
    for (long long j = 1; j <= k; j++)
      for (long long p = 1; p <= k; p++)
        c.g[i][j] = add(c.g[i][j], mul(a.g[i][p], b.g[p][j], Mod - 1), Mod - 1);
  return c;
}
Matrix fast_pow(Matrix a, long long b) {
  Matrix res;
  for (long long i = 1; i <= k; i++) res.g[i][i] = 1;
  for (; b; b >>= 1, a = a * a)
    if (b & 1) res = res * a;
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return;
  }
  exgcd(b, a % b, y, x);
  y -= a / b * x;
}
long long exgcd(long long a, long long b, long long c) {
  long long g = gcd(a, b);
  if (c % g) return -1;
  long long x, y;
  exgcd(a, b, x, y);
  x *= c / g;
  x = (x % (b / g) + (b / g)) % (b / g);
  return x;
}
long long bsgs(long long a, long long b) {
  map<long long, long long> mp;
  mp[b] = 0;
  long long cur = 1, limit = sqrt(Mod);
  for (long long i = 1; i <= limit; i++) {
    cur = mul(cur, a);
    mp[mul(b, exgcd(cur, Mod, 1))] = i;
  }
  long long now = 1;
  for (long long i = 0; i <= limit; i++) {
    if (mp.count(now)) return limit * i + mp[now];
    now = mul(now, cur);
  }
  return -1;
}
int main() {
  scanf("%lld", &k);
  for (long long i = 1; i <= k; i++) scanf("%lld", &b[i]);
  scanf("%lld %lld", &n, &m);
  Matrix tmp;
  for (long long i = 1; i < k; i++) tmp.g[i][i + 1] = 1;
  for (long long i = 1; i <= k; i++) tmp.g[k][i] = b[k - i + 1];
  tmp = fast_pow(tmp, n - k);
  long long ans1 = bsgs(3, m), ans2 = exgcd(tmp.g[k][k], Mod - 1, ans1);
  if (ans1 == -1 || ans2 == -1)
    printf("-1");
  else
    printf("%lld", fast_pow(3, ans2));
  return 0;
}