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>
namespace TNT {
const int mod = 998244353;
const int MAXN = 110;
int k, n, m;
inline int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
struct array {
  int ar[MAXN][MAXN];
  int x, y;
} a, asum, s;
array operator*(array x, array y) {
  array c;
  c.x = x.x;
  c.y = y.y;
  for (register int i = 1; i <= x.x; i++)
    for (register int j = 1; j <= y.y; j++) {
      c.ar[i][j] = 0;
      for (register int z = 1; z <= x.y; z++)
        c.ar[i][j] = 1ll * (1ll * c.ar[i][j] + 1ll * x.ar[i][z] * y.ar[z][j]) %
                     (mod - 1);
    }
  return c;
}
inline array qpow(array x, int y) {
  array ans = x;
  y--;
  while (y) {
    if (y & 1) ans = ans * x;
    x = x * x;
    y >>= 1;
  }
  return ans;
}
inline int power(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = 1ll * ans * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return ans;
}
int exgcd(int a, int b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  int d = exgcd(b, a % b, x, y);
  int g = x;
  x = y;
  y = g - a / b * y;
  return d;
}
std::map<int, int> hs;
inline int BSGS(int G) {
  m %= mod;
  int t = sqrt(mod) + 1, tmp = m % mod;
  hs[tmp] = 0;
  for (int i = 1; i < t; i++) {
    tmp = 1ll * tmp * G % mod;
    hs[tmp] = i;
  }
  G = power(G, t);
  if (!G) {
    if (!m)
      return 1;
    else
      return -1;
  }
  tmp = 1;
  if (hs.find(tmp) != hs.end()) return 0;
  for (int i = 1; i <= t; i++) {
    tmp = 1ll * tmp * G % mod;
    if (hs.find(tmp) != hs.end()) return i * t - hs[tmp];
  }
  return -1;
}
inline void main() {
  k = read();
  s.x = 1;
  a.x = a.y = s.y = k;
  s.ar[1][k] = 1;
  for (register int i = 2; i <= k; i++) a.ar[i][i - 1] = 1;
  for (register int i = k; i >= 1; i--) a.ar[i][k] = read();
  n = read(), m = read();
  asum = s * qpow(a, n - k);
  int t = BSGS(3);
  long long x, y;
  int d = exgcd(asum.ar[1][k], mod - 1, x, y);
  if (t % d)
    puts("-1");
  else {
    x = (t / d * x % (mod - 1) + mod - 1) % (mod - 1);
    printf("%d\n", power(3, x));
  }
  return;
}
}  // namespace TNT
int main() {
  TNT::main();
  return 0;
}