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;
template <typename T>
void mul(T &a, T &b, int k) {
  T c(k * k, 0);
  for (int i = 0; i < k; ++i)
    for (int j = 0; j < k; ++j) {
      for (int t = 0; t < k; ++t)
        c[i * k + j] =
            (c[i * k + j] + a[i * k + t] * b[t * k + j]) % (998244353 - 1);
    }
  copy(c.begin(), c.end(), a.begin());
}
template <typename T>
int64_t power(T &a, int k, int n) {
  T r(k * k);
  for (int i = 0; i < k; ++i) r[i * k + i] = 1;
  for (int t = 1; t <= n; t *= 2) {
    if (t & n) mul(r, a, k);
    mul(a, a, k);
  }
  return r[0];
}
int64_t power(int64_t x, int n) {
  int64_t r = 1;
  int t = 1;
  for (int t = 1; t <= n; t *= 2) {
    if (t & n) r = r * x % 998244353;
    x = x * x % 998244353;
  }
  return r;
}
template <typename T>
int find(T &u, int x) {
  int l = 0, r = u.size() - 1;
  while (l < r) {
    int i = (l + r) / 2;
    if (u[i].first < x) {
      l = i + 1;
    } else {
      r = i;
    }
  }
  if (l < u.size() && x == u[l].first)
    return u[l].second;
  else
    return -1;
}
int dlog(int64_t a, int64_t y) {
  vector<pair<int64_t, int>> u(31596), v(31596);
  int64_t x = power(a, 31596);
  int64_t r;
  int i, j;
  r = x;
  for (i = 0; i < 31596; ++i, r = r * x % 998244353)
    u[i] = pair<int64_t, int>(r, i);
  sort(u.begin(), u.end());
  r = y * a % 998244353;
  for (i = 0; i < 31596; ++i, r = r * a % 998244353)
    if ((j = find(u, r)) != -1) break;
  i++;
  j++;
  return (998244353 - 1 + 31596 * j - i) % (998244353 - 1);
}
int64_t gcd(int64_t a, int64_t b) {
  while (a % b) {
    int64_t c = a % b;
    a = b;
    b = c;
  }
  return b;
}
int64_t inverse(int64_t a, int64_t n) {
  int64_t tmp;
  int64_t t = 0, t1 = 1;
  int64_t r = n, r1 = a;
  while (r1 != 0) {
    int64_t q = r / r1;
    tmp = t - q * t1;
    t = t1;
    t1 = tmp;
    tmp = r - q * r1;
    r = r1;
    r1 = tmp;
  }
  if (t < 0) t = t % n + n;
  return t;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int n, m, k;
  cin >> k;
  vector<int64_t> a(k * k, 0);
  for (int i = 0; i < k; ++i) cin >> a[i];
  for (int i = 0; i < k - 1; ++i) a[k * (i + 1) + i] = 1;
  cin >> n >> m;
  int64_t e = power(a, k, n - k);
  int64_t y = dlog(3, m);
  if (e != 0) {
    int64_t d = gcd(e, 998244353 - 1);
    if (y % d != 0) {
      cout << -1 << endl;
    } else {
      int r = y / d * inverse(e / d, (998244353 - 1) / d) % (998244353 - 1);
      cout << power(3, r) << endl;
    }
  } else {
    if (y != 0) {
      cout << -1 << endl;
    } else {
      cout << 1 << endl;
    }
  }
  return 0;
}