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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> p) {
  os << '{' << p.first << ", " << p.second << '}';
  return os;
}
template <typename T>
inline void showArray(T *it1, T *it2) {
  cout << "{ ";
  while (it1 != it2) {
    cout << *it1 << ' ';
    ++it1;
  }
  cout << " }" << endl;
}
inline int readInt() {
  int n;
  char c = getchar(), k = '+';
  if (c == '-') {
    k = '-';
  } else {
    n = c - '0';
  }
  c = getchar();
  while (c != ' ' && c != '\n') {
    n = n * 10 + c - '0';
    c = getchar();
  }
  if (k == '-') {
    n = -n;
  }
  return n;
}
inline void printInt(int n) {
  if (n < 0) {
    putchar('-');
    n = -n;
  }
  char st[10], *p;
  p = st + 10;
  while (n) {
    *(--p) = (n % 10) + '0';
    n /= 10;
  }
  if (p == st + 10) {
    *(--p) = '0';
  }
  while (p != st + 10) {
    putchar(*p++);
  }
}
const bool debug = false;
const int MAXN = 1e3 + 2, INF = 1e9, p = 998244353, g = 3;
const double pi = acos(-1), eps = 1e-9;
int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
vector<vector<int>> operator*(const vector<vector<int>> &a,
                              const vector<vector<int>> &b) {
  vector<vector<int>> c(a.size(), vector<int>(b[0].size(), 0));
  for (size_t i = 0; i < a.size(); ++i) {
    for (size_t j = 0; j < b[0].size(); ++j) {
      for (size_t k = 0; k < b.size(); ++k) {
        c[i][j] = (c[i][j] + (long long)a[i][k] * b[k][j]) % (p - 1);
      }
    }
  }
  return c;
}
int fast_pow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) {
      res = ((long long)res * a) % p;
    }
    a = ((long long)a * a) % p;
    b >>= 1;
  }
  return res;
}
int fast_pow(int a, int b, int m) {
  int res = 1;
  while (b) {
    if (b & 1) {
      res = ((long long)res * a) % m;
    }
    a = ((long long)a * a) % m;
    b >>= 1;
  }
  return res;
}
vector<vector<int>> fast_pow(vector<vector<int>> a, int b) {
  vector<vector<int>> res(a.size(), vector<int>(a.size(), 0));
  for (size_t i = 0; i < a.size(); ++i) res[i][i] = 1;
  while (b) {
    if (b & 1) {
      res = res * a;
    }
    a = a * a;
    b >>= 1;
  }
  return res;
}
int disc_log(int a, int b) {
  int r = sqrt(p) + 1;
  map<int, int> value;
  int an = fast_pow(a, r);
  for (int i = 1, cur = an; i <= r; ++i) {
    if (value.find(cur) == value.end()) {
      value[cur] = i;
    }
    cur = ((long long)cur * an) % p;
  }
  for (int i = 0, cur = b; i <= r; ++i) {
    if (value.find(cur) != value.end()) {
      int ans = value[cur] * r - i;
      if (ans < p) return ans;
    }
    cur = ((long long)cur * a) % p;
  }
  return -1;
}
int f(int n) {
  int res = n;
  for (int i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      while (n % i == 0) {
        n /= i;
      }
      res -= res / i;
    }
  }
  if (n > 1) {
    res -= res / n;
  }
  return res;
}
int solve_mod_eq(int a, int b) {
  int d = gcd(a, p - 1);
  if (b % d) return -1;
  a /= d;
  b /= d;
  int m = (p - 1) / d;
  return ((long long)b * fast_pow(a, f(m) - 1, m)) % m;
}
int k, n, fn;
vector<vector<int>> A;
int main() {
  scanf("%d", &k);
  A.resize(k, vector<int>(k, 0));
  for (int i = 0; i < k; ++i) {
    scanf("%d", &A[0][i]);
    if (i) A[i][i - 1] = 1;
  }
  if (debug) {
    printf("A:\n");
    for (int i = 0; i < k; ++i) {
      for (int j = 0; j < k; ++j) {
        printf("%d ", A[i][j]);
      }
      printf("\n");
    }
  }
  scanf("%d %d", &n, &fn);
  int hn = disc_log(g, fn) % (p - 1);
  if (debug) {
    printf("hn = %d\n", hn);
  }
  if (hn == -1) {
    printf("-1\n");
    return 0;
  }
  A = fast_pow(A, n - k);
  if (debug) {
    printf("A ** n - k:\n");
    for (int i = 0; i < k; ++i) {
      for (int j = 0; j < k; ++j) {
        printf("%d ", A[i][j]);
      }
      printf("\n");
    }
  }
  int hk = solve_mod_eq(A[0][0], hn);
  if (hk == -1) {
    printf("-1\n");
    return 0;
  }
  hk = (hk + p - 1) % (p - 1);
  if (debug) {
    printf("hk = %d\n", hk);
  }
  printf("%d\n", fast_pow(g, hk));
  return 0;
}