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;
using ll = long long;
using ld = long double;
template <class T>
using vec = vector<T>;
template <class T>
using v2d = vector<vector<T>>;
template <class T>
using v3d = vector<vector<vector<T>>>;
template <class T>
using v4d = vector<vector<vector<vector<T>>>>;
template <class T>
using prq = priority_queue<T>;
template <class T>
using rpq = priority_queue<T, vector<T>, greater<T>>;
using Mat = v2d<ll>;
using Row = vec<ll>;
const int MOD = 998244353;
const int MOD_E = MOD - 1;
const int ROOT = 3;
const Mat ID = {{-1}};
Mat operator*(const Mat& a, const Mat& b) {
  if (b == ID) {
    return a;
  }
  assert(a[0].size() == b.size());
  int y = a.size(), x = b[0].size(), n = a[0].size();
  Mat c(y, Row(x));
  for (int i = 0; i < y; ++i) {
    for (int k = 0; k < n; ++k) {
      for (int j = 0; j < x; ++j) {
        c[i][j] = (c[i][j] + (a[i][k] * b[k][j]) % MOD_E) % MOD_E;
      }
    }
  }
  return c;
}
Mat pow(const Mat& x, int expo) {
  if (expo == 0) {
    return ID;
  }
  if (expo & 1) {
    return x * pow(x, expo - 1);
  }
  Mat temp = pow(x, expo / 2);
  return temp * temp;
}
ll mpow(ll n, ll k) {
  if (k == 0) {
    return 1;
  }
  if (k & 1) {
    return (n * mpow(n, k - 1)) % MOD;
  }
  ll temp = mpow(n, k / 2);
  return (temp * temp) % MOD;
}
int bsgs(int a, int b) {
  const int N = sqrt(MOD);
  map<int, int> res;
  for (int i = 1; i <= N; ++i) {
    res[mpow(a, N * i)] = i;
  }
  for (int i = 0; i <= N; ++i) {
    int cur = (b * mpow(a, i)) % MOD;
    auto ptr = res.find(cur);
    if (ptr != res.end()) {
      return ptr->second * N - i;
    }
  }
  return -1;
}
int solve(int k, int a) {
  int res = bsgs(mpow(ROOT, k), a);
  if (res == -1) {
    return -1;
  }
  return mpow(ROOT, res);
}
int main() {
  int k;
  scanf("%d", &k);
  vec<int> b(k);
  for (auto& i : b) {
    scanf("%d", &i);
  }
  int n, m;
  scanf("%d %d", &n, &m);
  Mat tr(k, Row(k));
  for (int i = 0; i < k; ++i) {
    tr[0][i] = b[i];
  }
  for (int i = 1; i < k; ++i) {
    tr[i][i - 1] = 1;
  }
  Mat base(1, Row(k));
  base[0][0] = 1;
  int times = n - k;
  auto res = base * pow(tr, times);
  printf("%d\n", solve(res[0][0], m));
}