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 N = 105, mod = 998244353, G = 3;
int K, A, n, m;
struct matrix {
  int a[N][N];
  matrix() { memset(a, 0, sizeof(a)); }
  matrix operator*(const matrix &x) const {
    matrix res;
    for (int i = 1; i <= K; i++)
      for (int j = 1; j <= K; j++)
        for (int k = 1; k <= K; k++)
          (res.a[i][j] += 1ll * a[i][k] * x.a[k][j] % (mod - 1)) %= (mod - 1);
    return res;
  }
  matrix mpow(int b) const {
    matrix res, bs = *this;
    for (int i = 1; i <= K; i++)
      for (int j = 1; j <= K; j++) bs.a[i][j] = a[i][j];
    for (int i = 1; i <= K; i++) res.a[i][i] = 1;
    for (; b; bs = bs * bs, b >>= 1)
      if (b & 1) res = res * bs;
    return res;
  }
};
matrix g, st;
map<int, int> mp;
int ksm(int a, int b, int p) {
  int x = 1;
  for (; b; b & 1 ? x = 1ll * x * a % p : 0, a = 1ll * a * a % p, b >>= 1)
    ;
  return x;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int exgcd(int a, int b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int d = exgcd(b, a % b, x, y);
  long long t = x;
  x = y;
  y = t - (a / b) * y;
  return d;
}
int inv(int a, int p) {
  long long x, y;
  exgcd(a, p, x, y);
  return (x % p + p) % p;
}
int bsgs(int a, int b, int p) {
  int sp = int(sqrt(p)) + 1, pw = ksm(a, sp, p);
  for (int i = 0, j = b; i <= sp; i++, j = 1ll * j * a % p)
    mp[j] = max(mp[j], i);
  for (int i = 1, j = pw; i <= sp; i++, j = 1ll * j * pw % p)
    if (mp.count(j)) return i * sp - mp[j];
  return -1;
}
int main() {
  int x;
  scanf("%d", &K);
  for (int i = 1; i <= K; i++) scanf("%d", &x), g.a[K - i + 1][K] = x;
  scanf("%d%d", &n, &m);
  for (int i = 2; i <= K; i++) g.a[i][i - 1] = 1;
  g = g.mpow(n - K);
  st.a[1][K] = 1;
  st = st * g;
  A = st.a[1][K];
  int gn = bsgs(G, m, mod), gc = gcd(A, mod - 1), p = (mod - 1) / gc;
  if (gn % gc) {
    puts("-1");
    return 0;
  }
  printf("%d\n", ksm(G, 1ll * (gn / gc) * inv(A / gc, p) % p, mod));
  return 0;
}