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 MK = 100;
const long long MOD = 998244353;
struct matrix {
  long long val[MK][MK];
  void clear() { memset(val, 0, sizeof(val)); }
  void m1() {
    clear();
    for (int i = 0; i < MK; ++i) {
      val[i][i] = 1;
    }
  }
  matrix(){};
};
matrix cross(const matrix &a, const matrix &b, long long MOD) {
  matrix res;
  res.clear();
  for (int i = 0; i < MK; ++i) {
    for (int j = 0; j < MK; ++j) {
      for (int k = 0; k < MK; ++k) {
        res.val[i][j] += a.val[i][k] * b.val[k][j];
        res.val[i][j] %= MOD;
      }
    }
  }
  return res;
}
matrix bipow(matrix a, long long pow, long long MOD) {
  if (pow == 1) return a;
  matrix res;
  res.m1();
  if (pow == 0) return res;
  while (pow) {
    if (pow & 1) res = cross(res, a, MOD);
    a = cross(a, a, MOD);
    pow >>= 1;
  }
  return res;
}
long long binup(long long a, long long pow, long long MOD) {
  if (pow == 0) return 1;
  if (pow & 1) return a * binup(a, pow - 1, MOD) % MOD;
  long long b = binup(a, pow / 2, MOD);
  return b * b % MOD;
}
long long powmod(long long a, long long b, long long m) {
  if (b == 1) return a;
  long long res = 1;
  while (b > 0)
    if (b & 1) {
      res = (res * a) % m;
      --b;
    } else {
      a = (a * a) % m;
      b >>= 1;
    }
  return res % m;
}
long long solve(long long a, long long b, long long m) {
  long long n = (long long)sqrt(m + .0) + 1;
  map<long long, long long> vals;
  for (long long i = n; i >= 1; --i) vals[powmod(a, i * n, m)] = i;
  for (long long i = 0; i <= n; ++i) {
    long long cur = (powmod(a, i, m) * b) % m;
    if (vals.count(cur)) {
      long long ans = vals[cur] * n - i;
      if (ans < m) return ans;
    }
  }
  return -1;
}
long long b[1000];
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int k;
  cin >> k;
  for (int i = 0; i < k; ++i) {
    cin >> b[i];
  }
  long long n, m;
  cin >> n >> m;
  matrix start;
  start.clear();
  for (int i = 0; i < k - 1; ++i) {
    start.val[0][i] = 0;
  }
  start.val[0][k - 1] = 1;
  matrix nxt;
  nxt.clear();
  for (int i = 0; i < k - 1; ++i) {
    nxt.val[i][i + 1] = 1;
  }
  for (int i = 0; i < k; ++i) {
    nxt.val[k - 1][i] = b[k - i - 1];
  }
  nxt = bipow(nxt, n - k, MOD - 1);
  start = cross(start, nxt, MOD - 1);
  long long pow = start.val[0][k - 1];
  long long g = 3;
  long long second = solve(binup(g, pow, MOD), m, MOD);
  if (second == -1) {
    cout << second;
  } else {
    cout << binup(g, second, MOD);
  }
}