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 long long MOD = 998244353LL;
const long long dx[4] = {-1, -1, 1, 1};
const long long dy[4] = {-1, 1, -1, 1};
const long long MAX = 2e5 + 50;
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long n, x, y, k, Q, ans, d;
string second;
long long res[102][102], ret[102][102], m[102][102], tmp[102][102];
void matrixMultiplyret() {
  for (int i(0); i < n; i++) {
    for (int j(0); j < n; j++) {
      tmp[i][j] = 0;
      for (int k(0); k < n; k++) {
        tmp[i][j] += ((ret[i][k] + (MOD - 1)) % (MOD - 1) *
                      (m[k][j] + (MOD - 1)) % (MOD - 1)) %
                     (MOD - 1);
        tmp[i][j] = (tmp[i][j] + (MOD - 1)) % (MOD - 1);
      }
    }
  }
  for (int i(0); i < n; i++)
    for (int j(0); j < n; j++) ret[i][j] = tmp[i][j];
}
void matrixMultiplymat() {
  for (int i(0); i < n; i++) {
    for (int j(0); j < n; j++) {
      tmp[i][j] = 0;
      for (int k(0); k < n; k++) {
        tmp[i][j] += ((m[i][k] + (MOD - 1)) % (MOD - 1) *
                      (m[k][j] + (MOD - 1)) % (MOD - 1)) %
                     (MOD - 1);
        tmp[i][j] = (tmp[i][j] + (MOD - 1)) % (MOD - 1);
      }
    }
  }
  for (int i(0); i < n; i++)
    for (int j(0); j < n; j++) m[i][j] = tmp[i][j];
}
void matrixPower(long long pow) {
  for (int i(0); i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j)
        ret[i][i] = 1;
      else
        ret[i][j] = 0;
    }
  }
  while (pow) {
    if (pow & 1) {
      matrixMultiplyret();
    }
    pow >>= 1;
    matrixMultiplymat();
  }
}
long long pw(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ret;
}
pair<long long, long long> solve(long long x, long long y) {
  if (y > x) {
    auto u = solve(y, x);
    swap(u.first, u.second);
    return u;
  }
  if (!y) return {1, 0};
  long long r = x / y;
  auto u = solve(y, x % y);
  long long a = u.first, b = u.second;
  return {b, (a - (r * b) % (MOD - 1)) % (MOD - 1)};
}
void process() {
  map<long long, long long> tmp;
  tmp[1] = 0;
  long long prev = 1LL;
  for (long j = 1; j < 32500LL; j++) {
    prev = (prev * 3LL) % MOD;
    tmp[prev] = j;
  }
  long long a = pw(pw(3, MOD - 2), 32500LL);
  long long cp = 1LL;
  for (long j = 0;; j++) {
    long long cur = (cp * x) % MOD;
    cp = (cp * a) % MOD;
    if (tmp.find(cur) != tmp.end()) {
      x = (tmp[cur] + 32500LL * j) % (MOD - 1);
      return;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++)
    for (int j = 0; j < n; j++) m[i][j] = ((i + 1) == j);
  for (int j = 0; j < n; j++) {
    cin >> m[n - 1][n - 1 - j];
    m[n - 1][n - 1 - j] %= (MOD - 1);
  }
  cin >> k >> x;
  process();
  matrixPower(k - n);
  long long last = ret[n - 1][n - 1];
  if (!last && !x) return cout << 1 << '\n', 0;
  if (!last) return cout << -1 << '\n', 0;
  long long tmp = (MOD - 1);
  long long g = gcd(tmp, last);
  if (x % g != 0) return cout << -1 << '\n', 0;
  last /= g;
  tmp /= g;
  x /= g;
  last %= tmp;
  ans = (solve(tmp, last).second * x) % tmp;
  ans = (ans + tmp) % tmp;
  cout << pw(3, ans) << '\n';
  return 0;
}