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 MXSZ = 1e5 + 5;
const long long mod = 998244353;
const long long m0d = 998244352;
struct Matrix {
  long long m[103][103];
  Matrix() { memset(m, 0, sizeof(m)); }
  Matrix operator*(Matrix const &rhs) {
    Matrix ret;
    for (int i = 1; i <= 100; i++) {
      for (int j = 1; j <= 100; j++) {
        ret.m[i][j] = 0;
        for (int il = 1; il <= 100; il++) {
          ret.m[i][j] += ((m[i][il] * rhs.m[il][j]) % m0d);
          ret.m[i][j] %= m0d;
        }
      }
    }
    return ret;
  }
  void print(int n) {
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) cout << m[i][j] << " \n"[j == n];
    cout << "================\n";
  }
};
long long n, m, k;
long long b[MXSZ];
Matrix base, ans;
void fpow(Matrix b, long long e) {
  Matrix &ret = ans;
  for (int i = 1; i <= k; i++) ret.m[i][i] = 1;
  while (e > 0) {
    if (e & 1) {
      ret = ret * b;
    }
    b = b * b;
    e >>= 1;
  }
}
long long fpow2(long long b, long long e) {
  long long ret = 1;
  while (e > 0) {
    if (e & 1) {
      (ret *= b) %= mod;
    }
    b *= b;
    b %= mod;
    e >>= 1;
  }
  return ret;
}
long long g = 3LL, primroot = 1;
map<long long, long long> mp;
long long BSGS() {
  long long l = 1, r = mod, sq = 1;
  while (l <= r) {
    long long md = (l + r) >> 1;
    if (md * md > mod) {
      r = md - 1;
    } else {
      l = md + 1;
      sq = md;
    }
  }
  for (int i = 1; i <= sq; i++) {
    mp[fpow2(primroot, i * sq)] = i;
  }
  for (int i = 0; i <= sq; i++) {
    long long cur = fpow2(primroot, i) * m;
    cur %= mod;
    if (mp.count(cur)) {
      long long ans = mp[cur] * sq - i;
      if (ans < mod) return ans;
    }
  }
  return -1;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> k;
  for (int i = 1; i <= k; i++) {
    cin >> b[i];
  }
  cin >> n >> m;
  for (int i = 1; i <= k; i++) {
    base.m[1][i] = b[i];
  }
  for (int i = 2; i <= k; i++) {
    base.m[i][i - 1] = 1;
  }
  fpow(base, n - k);
  long long x = ans.m[1][1];
  primroot = fpow2(g, x);
  long long POWER = BSGS();
  if (POWER == -1) {
    cout << -1 << '\n';
    return 0;
  }
  cout << fpow2(g, POWER) << '\n';
  return 0;
}