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;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
const long long mod = 998244353, g = 3;
long long qpow(long long a, int k, const long long m = mod) {
  long long ans = 1;
  while (k) {
    if (k & 1) (ans *= a) %= m;
    (a *= a) %= m;
    k >>= 1;
  }
  return ans;
}
long long generator(int p) {
  vector<int> fact;
  int phi = p - 1, n = p - 1;
  for (int i = 2; i * i <= p; ++i) {
    if (n % i == 0) {
      fact.push_back(i);
      while (n % i == 0) {
        n /= i;
      }
    }
  }
  if (n > 1) fact.push_back(n);
  for (int ans = 2; ans <= p; ++ans) {
    bool ok = true;
    for (auto it : fact) {
      ok &= pow(ans, phi / it) != 1;
      if (!ok) break;
    }
    if (ok) return ans;
  }
  return -1;
}
int K;
struct matrix {
  long long m[100][100];
  matrix() { memset(m, 0, sizeof m); }
  matrix(matrix &other) { memcpy(m, other.m, sizeof m); }
  long long *operator[](int n) { return m[n]; }
  matrix operator*(matrix &other) {
    matrix ans;
    for (int i = 0; i < K; ++i) {
      for (int k = 0; k < K; ++k) {
        for (int j = 0; j < K; ++j) {
          (ans[i][j] += m[i][k] * other[k][j]) %= (mod - 1);
        }
      }
    }
    return ans;
  }
  void pow(int k) {
    --k;
    matrix a(*this);
    while (k) {
      if (k & 1) (*this) = a * (*this);
      a = a * a;
      k >>= 1;
    }
  }
  void print() const {
    for (int i = 0; i < K; ++i, putchar('\n')) {
      for (int j = 0; j < K; ++j) {
        printf("%4lld", m[i][j]);
      }
    }
  }
} mat;
long long gcd(long long x, long long y) {
  long long temp;
  while (y) {
    temp = x % y;
    x = y;
    y = temp;
  }
  return x;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  long long q = exgcd(b, a % b, y, x);
  y -= a / b * x;
  return q;
}
long long phi(long long x) {
  long long ans = x;
  for (int i = 2; i * i <= x; ++i) {
    if (x % i == 0) {
      ans = ans / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) ans = ans / x * (x - 1);
  return ans;
}
long long get_inverse(long long a, long long c, long long m) {
  long long p = gcd(a, m);
  if (c % p != 0) return -1;
  a /= p, c /= p;
  long long new_mod = m / p;
  return qpow(a, phi(new_mod) - 1, new_mod) * c % new_mod;
}
long long BSGS(long long a, long long b, const long long m) {
  long long n = sqrt(m) + 1;
  long long cur = b;
  unordered_map<long long, long long, custom_hash> val;
  for (long long p = 0; p < n; ++p) {
    val[cur] = p;
    cur = (cur * g) % m;
  }
  cur = qpow(g, n, m);
  long long now = cur;
  for (long long q = 1; q <= n + 1; ++q) {
    if (val.count(now)) return q * n - val[now];
    now = now * cur % m;
  }
  return -1;
}
int main() {
  scanf("%d", &K);
  for (int i = 0; i < K; ++i) scanf("%lld", mat[0] + i), mat[0][i] %= (mod - 1);
  for (int i = 1; i < K; ++i) mat[i][i - 1] = 1;
  int n, m;
  scanf("%d%d", &n, &m);
  mat.pow(n - K);
  long long c = mat[0][0];
  long long pm = BSGS(3, m, mod);
  long long px = get_inverse(c, pm, mod - 1);
  if (px == -1) {
    printf("-1\n");
  } else {
    printf("%lld\n", qpow(3, px, mod));
  }
  return 0;
}