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 maxk = 110;
const int mod = 998244353;
int n, m, K;
int a[maxk];
inline int read() {
  char c = getchar();
  int res = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
  return res * f;
}
struct Matrix {
  int n, m;
  int a[maxk][maxk];
  Matrix() {
    memset(a, 0, sizeof(a));
    n = m = 0;
  }
  int* operator[](const int i) { return a[i]; }
  inline void idx() {
    for (int i = 1; i <= min(n, m); i++) a[i][i] = 1;
  }
} A, B;
Matrix operator*(Matrix a, Matrix b) {
  Matrix res;
  res.n = a.n, res.m = b.m;
  for (int i = 1; i <= a.n; i++)
    for (int j = 1; j <= b.m; j++)
      for (int k = 1; k <= res.m; k++)
        res[i][j] =
            (res[i][j] + 1ll * a[i][k] * b[k][j] % (mod - 1)) % (mod - 1);
  return res;
}
Matrix operator^(Matrix x, int k) {
  Matrix res;
  res.n = x.n, res.m = x.m;
  res.idx();
  while (k) {
    if (k & 1) res = res * x;
    x = x * x;
    k >>= 1;
  }
  return res;
}
inline int power(int x, int k, int mod) {
  int res = 1;
  while (k) {
    if (k & 1) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod;
    k >>= 1;
  }
  return res;
}
unordered_map<int, int> mp;
inline int BSGS(int a, int b, int mod) {
  if (!b) return 0;
  if (!a) return !b ? 1 : -1;
  int t = sqrt(mod + 0.5);
  for (int i = 0, now = 1; i <= t; i++, now = 1ll * now * a % mod)
    mp[1ll * b * now % mod] = i;
  a = power(a, t, mod);
  for (int i = 1, now = a; i <= t; i++, now = 1ll * now * a % mod)
    if (mp.count(now)) return i * t - mp[now];
  return -1;
}
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 d = exgcd(b, a % b, x, y);
  long long z = x;
  x = y, y = z - (a / b) * y;
  return d;
}
int main() {
  K = read();
  for (int i = 1; i <= K; i++) a[i] = read();
  n = read(), m = read();
  A.n = 1, A.m = K;
  A[1][K] = 1;
  B.n = B.m = K;
  for (int i = 1; i <= K - 1; i++) B[i + 1][i] = 1;
  for (int i = 1; i <= K; i++) B[i][K] = a[K - i + 1];
  B = B ^ (n - K);
  A = A * B;
  int t = BSGS(3, m, mod);
  long long x, y, d;
  d = exgcd(A[1][K], mod - 1, x, y);
  if (t % d) {
    puts("-1");
    return 0;
  }
  x = (1ll * t / d * x % (mod - 1) + (mod - 1)) % (mod - 1);
  printf("%d", power(3, x, mod));
  return 0;
}