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;
template <typename T>
void read(T &t) {
  t = 0;
  char ch = getchar();
  int f = 1;
  while ('0' > ch || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  do {
    (t *= 10) += ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  t *= f;
}
const long long mod = 998244353;
int k, n, m, A, B, g = 3;
map<int, int> M;
struct Matrix {
  long long d[110][110];
  int n, m;
  void init() {
    n = m = k;
    memset(d, 0, sizeof(d));
    for (int i = 1; i <= n; i++) d[i][i] = 1;
  }
} a, b, ans;
Matrix cheng(Matrix x, Matrix y) {
  Matrix z;
  z.n = x.n;
  z.m = y.m;
  for (int i = 1; i <= z.n; i++)
    for (int j = 1; j <= z.m; j++) {
      z.d[i][j] = 0;
      for (int K = 1; K <= x.m; K++)
        z.d[i][j] += x.d[i][K] * y.d[K][j] % (mod - 1);
      z.d[i][j] %= (mod - 1);
    }
  return z;
}
long long ksm(long long x, long long y, long long p) {
  if (y == 0) return 1;
  if (y == 1) return x % p;
  long long res = ksm(x, y / 2, p);
  res = res * res % p;
  if (y & 1) res = res * x % p;
  return res;
}
long long gcd(long long x, long long y) {
  if (!x || !y) return x + y;
  return gcd(y, x % y);
}
void exgcd(int a, int b, int &x, int &y) {
  if (b == 0)
    x = 1, y = 0;
  else {
    exgcd(b, a % b, y, x);
    y -= a / b * x;
  }
}
int inverse(int a, int m) {
  int x, y;
  exgcd(a, m, x, y);
  return (x % m + m) % m;
}
int main() {
  read(k);
  for (int i = 1; i <= k; i++) read(b.d[k - i + 1][k]);
  read(n);
  read(B);
  for (int i = 1; i < k; i++) b.d[i + 1][i] = 1;
  b.n = b.m = k;
  ans.init();
  n -= k;
  while (n) {
    if (n & 1) ans = cheng(ans, b);
    n /= 2;
    b = cheng(b, b);
  }
  a.d[1][k] = 1;
  a.n = 1;
  a.m = k;
  ans = cheng(a, ans);
  A = ans.d[1][k];
  m = ceil(sqrt(mod));
  long long tmp = B % mod;
  for (int i = 1; i <= m; i++) {
    tmp = tmp * g % mod;
    M[tmp] = i;
  }
  long long iv = ksm(g, m, mod);
  tmp = 1;
  int indb, d = gcd(A, mod - 1);
  for (int i = 1; i <= m; i++) {
    tmp = tmp * iv % mod;
    if (M[tmp]) {
      indb = i * m - M[tmp];
      break;
    }
  }
  if (indb % d != 0) {
    printf("-1\n");
    return 0;
  }
  indb /= d;
  indb = (long long)indb * inverse(A / d, (mod - 1) / d) % ((mod - 1) / d);
  printf("%lld\n", ksm(g, indb, mod));
  return 0;
}