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 P = 998244353, Q = P - 1, R = 3;
const int N = 110;
int k;
struct Mat {
  int g[N][N];
  Mat() { memset(g, 0, sizeof(g)); }
  Mat(const Mat& rhs) { memcpy(g, rhs.g, sizeof(g)); }
  int* operator[](int ind) { return g[ind]; }
  const int* operator[](int ind) const { return g[ind]; }
  Mat operator*(const Mat& rhs) const {
    Mat ret = Mat();
    for (int i = 1; i <= k; ++i)
      for (int j = 1; j <= k; ++j)
        for (int l = 1; l <= k; ++l)
          ret[i][j] = (ret[i][j] + g[i][l] * (long long)rhs[l][j]) % Q;
    return ret;
  }
};
int b;
unordered_map<long long, long long> giant;
int mpow(long long a, long long k, int m) {
  long long ret = 1;
  while (k) {
    if (k & 1) ret = ret * (long long)a % m;
    a = a * (long long)a % m;
    k >>= 1;
  }
  return ret;
}
int ggcd;
void exGcd(int a, int b, int& x, int& y) {
  if (!b) {
    ggcd = a;
    x = 1;
    y = 0;
    return;
  }
  exGcd(b, a % b, y, x);
  y -= a / b * x;
}
int ford(int v) {
  b = sqrt(Q);
  long long step = mpow(R, b, P), prod = 1;
  for (int i = 0; i * (long long)b < Q; ++i) {
    unordered_map<long long, long long>::iterator it = giant.find(prod);
    if (it != giant.end()) break;
    giant[prod] = i * (long long)b;
    prod = prod * (long long)step % P;
  }
  prod = v;
  int ans = Q;
  unordered_map<long long, long long>::iterator it;
  for (int i = 0; i < b; ++i) {
    it = giant.find(prod);
    if (it != giant.end()) {
      ans = it->second - i;
      break;
    }
    prod = prod * (long long)R % P;
  }
  return ans;
}
int main() {
  scanf("%d", &k);
  Mat gen = Mat();
  for (int i = 1; i < k; ++i) gen[i][i + 1] = 1;
  for (int i = 1; i <= k; ++i) scanf("%d", &gen[i][1]);
  Mat res = Mat();
  for (int i = 1; i <= k; ++i) res[i][i] = 1;
  int n, m;
  scanf("%d%d", &n, &m);
  n -= 1;
  while (n) {
    if (n & 1) res = res * gen;
    if (n >>= 1) gen = gen * gen;
  }
  int ord = res[1][k];
  int ind = ford(m);
  int x, y;
  exGcd(ord, Q, x, y);
  int inv = x;
  if (inv < 0) inv += Q;
  if (ind % ggcd != 0)
    puts("-1");
  else
    printf("%d\n", mpow(R, (ind / ggcd) * (long long)inv % Q, P));
  return 0;
}