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 MOD = 998244353, g = 3, size = sqrt(MOD - 1);
long long n, k, m;
struct Matrix {
  long long a[105][105];
  Matrix() { memset(a, 0, sizeof(a)); }
  Matrix operator*(Matrix b) {
    Matrix ans;
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= k; j++) {
        for (int l = 1; l <= k; l++)
          ans.a[i][j] += a[i][l] * b.a[l][j] % (MOD - 1);
        ans.a[i][j] %= (MOD - 1);
      }
    return ans;
  }
} A;
long long s[100005], t[100005];
map<long long, int> mp;
long long quick_powm(Matrix x, long long a) {
  Matrix ans;
  for (int i = 1; i <= k; i++) ans.a[i][i] = 1;
  while (a) {
    if (a & 1) ans = ans * x;
    x = x * x;
    a >>= 1;
  }
  return ans.a[1][1];
}
long long quick_pow(long long x, long long a, long long MOD) {
  long long ans = 1;
  while (a) {
    if (a & 1) ans = ans * x % MOD;
    x = x * x % MOD;
    a >>= 1;
  }
  return ans;
}
void pre() {
  t[0] = 1;
  mp[1] = 0;
  for (int i = 1; i <= size; i++) {
    t[i] = t[i - 1] * g % MOD;
    mp[t[i]] = i;
  }
  for (int i = 0; i <= size; i++) s[i] = quick_pow(g, i * size, MOD);
}
long long ask(long long x) {
  for (int i = 0; i <= size; i++) {
    long long rest = x * quick_pow(s[i], MOD - 2, MOD) % MOD;
    if (rest == 1) return i * size;
    if (mp[rest] != 0) return mp[rest] + i * size;
  }
  return 0;
}
void exgcd(int x, int y, long long& a, long long& b, int d) {
  if (x > y) return exgcd(y, x, b, a, d);
  if (x == 0) {
    if (d % y == 0) {
      b = d / y;
      a = 0;
    }
    return;
  }
  exgcd(y % x, x, b, a, d);
  a -= b * (y / x);
}
int main() {
  scanf("%I64d", &k);
  for (int i = 1; i <= k; i++) scanf("%I64d", &A.a[1][i]);
  for (int i = 2; i <= k; i++) A.a[i][i - 1] = 1;
  scanf("%I64d%I64d", &n, &m);
  n -= k;
  long long x = quick_powm(A, n);
  pre();
  m = ask(m);
  long long a = 0, b = 0;
  exgcd(x, MOD - 1, a, b, m);
  a %= (MOD - 1);
  if (a < 0) a += MOD - 1;
  if (a * x % (MOD - 1) == m)
    printf("%I64d\n", quick_pow(g, a, MOD));
  else
    printf("-1\n");
  return 0;
}