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;
long long mod = (998244352LL);
inline long long Mod(long long a, long long b) { return (a % b); }
inline long long poww(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b, swap(a, b);
  }
  return a;
}
void read(vector<int>& w, int n) {
  w.resize(n);
  for (int i = 0; i < n; i++) cin >> w[i];
}
void print(vector<int>& w) {
  for (int i = 0; i < (int)(w).size(); i++) {
    if (i == (int)(w).size() - 1)
      cout << w[i] << "\n";
    else
      cout << w[i] << " ";
  }
}
int prodmod(vector<int> w);
int summod(vector<int> w);
int n, m, q, k, v[120], ans;
pair<int, int> w[120];
string second;
int b[120];
struct DiscreteRoot {
  inline int generator(int p) {
    vector<int> fact;
    int phi = p - 1, n = phi;
    for (int i = 2; i * i <= n; ++i) {
      if (n % i == 0) {
        fact.push_back(i);
        while (n % i == 0) n /= i;
      }
    }
    if (n > 1) fact.push_back(n);
    for (int res = 2; res <= p; ++res) {
      bool ok = true;
      for (int factor : fact) {
        if (poww(res, phi / factor) == 1) {
          ok = false;
          break;
        }
      }
      if (ok) return res;
    }
    return -1;
  }
  inline void solve(long long n, long long k, long long a) {
    mod = n;
    long long g = generator(n);
    long long sq = (long long)sqrt(n + .0) + 1;
    vector<pair<long long, long long> > dec(sq);
    for (long long i = 1; i <= sq; ++i)
      dec[i - 1] = {poww(1LL * g, 1LL * i * sq * k % (n - 1)), i};
    sort(dec.begin(), dec.end());
    long long any_ans = -1;
    for (long long i = 0; i < sq; ++i) {
      long long my = poww(g, i * k % (n - 1)) * a % n;
      auto it = lower_bound(dec.begin(), dec.end(),
                            pair<long long, long long>(my, 0));
      if (it != dec.end() && it->first == my) {
        any_ans = it->second * sq - i;
        break;
      }
    }
    if (any_ans == -1) {
      puts("-1");
      exit(0);
    } else {
      long long delta = (n - 1) / gcd(k, n - 1);
      for (long long cur = any_ans % delta; cur < n - 1; cur += delta) {
        cout << poww(1LL * g, 1LL * cur) << "\n";
        exit(0);
      }
    }
  }
};
template <class T, int N>
struct Matrix {
  array<array<T, N>, N> d{};
  Matrix operator*(const Matrix& m) const {
    Matrix a;
    for (int i = (0); i < (N); ++i)
      for (int j = (0); j < (N); ++j)
        for (int k = (0); k < (N); ++k) {
          a.d[i][j] += d[i][k] * m.d[k][j];
          a.d[i][j] %= mod;
        }
    return a;
  }
  vector<T> operator*(const vector<T>& vec) const {
    vector<T> ret(N);
    for (int i = (0); i < (N); ++i)
      for (int j = (0); j < (N); ++j) {
        ret[i] += d[i][j] * vec[j];
        ret[i] %= mod;
      }
    return ret;
  }
  Matrix operator^(long long p) const {
    assert(p >= 0);
    Matrix a, b(*this);
    for (int i = (0); i < (N); ++i) a.d[i][i] = 1;
    while (p) {
      if (p & 1) a = a * b;
      b = b * b;
      p >>= 1;
    }
    return a;
  }
};
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> k;
  for (int i = 1; i <= k; i++) cin >> b[i];
  cin >> n >> m;
  Matrix<long long, 100> base;
  for (int j = 0; j < k - 1; j++) {
    base.d[j + 1][j] = 1;
  }
  for (int i = 0; i < k; i++) base.d[i][k - 1] = b[k - i];
  Matrix<long long, 100> Matrix;
  Matrix.d[0][k - 1] = 1;
  base = (base ^ (n - 1));
  Matrix = Matrix * base;
  int alpha = Matrix.d[0][0];
  mod++;
  DiscreteRoot R;
  R.solve(mod, alpha, m);
}
int summod(vector<int> w) {
  int curr = 0;
  for (int i = 0; i < (int)(w).size(); i++) {
    curr = (curr + w[i]) % mod;
    if (curr < 0) curr += mod;
  }
  return curr;
}
int prodmod(vector<int> w) {
  int curr = 1;
  for (int i = 0; i < (int)(w).size(); i++) {
    if (w[i] >= mod) w[i] %= mod;
    curr = (curr * w[i]) % mod;
    if (curr < 0) curr += mod;
  }
  return curr;
}