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 INF = 0x3f3f3f3f;
const int MAXN = 500000 + 5;
const int MAXM = 600000 + 5;
const int SIZE = 101;
const long long MOD = 998244353;
struct Matrix {
  int r, c;
  long long mat[SIZE][SIZE];
  Matrix() {}
  Matrix(int _r, int _c) {
    r = _r;
    c = _c;
    memset(mat, 0, sizeof(mat));
  }
};
Matrix operator*(Matrix a, Matrix b) {
  Matrix s(a.r, b.c);
  for (int i = 0; i < a.r; i++) {
    for (int k = 0; k < a.c; k++) {
      for (int j = 0; j < b.c; j++) {
        s.mat[i][j] = (s.mat[i][j] + a.mat[i][k] * b.mat[k][j]) % (MOD - 1);
      }
    }
  }
  return s;
}
Matrix pow_mod(Matrix a, long long n) {
  Matrix ret(a.r, a.c);
  for (int i = 0; i < a.r; i++) {
    ret.mat[i][i] = 1;
  }
  Matrix tmp(a);
  while (n) {
    if (n & 1) ret = ret * tmp;
    tmp = tmp * tmp;
    n >>= 1;
  }
  return ret;
}
long long mul(long long a, long long b, long long m) {
  long long ret = 0;
  a %= m;
  while (b) {
    if (b & 1) ret = (ret + a) % m;
    a = (a + a) % m;
    b >>= 1;
  }
  return ret;
}
long long a_b_MOD_c(long long a, long long b, long long m) {
  long long ret = 1;
  a %= m;
  while (b) {
    if (b & 1) ret = mul(ret, a, m);
    a = mul(a, a, m);
    b >>= 1;
  }
  return ret;
}
long long ext_gcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long ret = ext_gcd(b, a % b, y, x);
  y -= a / b * x;
  return ret;
}
vector<long long> a;
bool g_test(long long g, long long p) {
  for (long long i = 0; i < a.size(); ++i)
    if (a_b_MOD_c(g, (p - 1) / a[i], p) == 1) return 0;
  return 1;
}
long long pri_root(long long p) {
  a.clear();
  long long tmp = p - 1;
  for (long long i = 2; i <= tmp / i; ++i)
    if (tmp % i == 0) {
      a.push_back(i);
      while (tmp % i == 0) tmp /= i;
    }
  if (tmp != 1) a.push_back(tmp);
  long long g = 1;
  while (true) {
    if (g_test(g, p)) return g;
    ++g;
  }
}
const int HASH_MOD = 9876543;
long long key[HASH_MOD], val[HASH_MOD];
int head[HASH_MOD], nxt[HASH_MOD];
struct Hash {
  int tot;
  void init() {
    memset(head, -1, sizeof(head));
    tot = 0;
  }
  long long insert(long long x, long long y) {
    int k = x % HASH_MOD;
    key[tot] = x;
    val[tot] = y;
    nxt[tot] = head[k];
    head[k] = tot++;
  }
  long long find(long long x) {
    int k = x % HASH_MOD;
    for (int i = head[k]; i != -1; i = nxt[i])
      if (key[i] == x) return val[i];
    return -1;
  }
} hs;
long long log_mod(long long a, long long b, long long m) {
  hs.init();
  long long s = ceil(sqrt(m + 0.5));
  long long cur = 1;
  for (int i = 0; i < s; ++i) {
    if (hs.find(cur) == -1) hs.insert(cur, i);
    cur = cur * a % m;
  }
  long long v = a_b_MOD_c(a, (m - s - 1 + m) % m, m);
  for (int i = 0; i < s; ++i) {
    long long tmp = hs.find(b);
    if (tmp != -1) return s * i + tmp;
    b = b * v % m;
  }
  return -1;
}
vector<long long> residue(long long p, long long N, long long a) {
  long long g = pri_root(p);
  g %= p;
  long long m = log_mod(g, a, p);
  vector<long long> ret;
  if (a == 0) {
    ret.push_back(0);
    return ret;
  }
  if (m == -1) return ret;
  long long A = N, B = p - 1, C = m, x, y;
  long long d = ext_gcd(A, B, x, y);
  if (C % d != 0) return ret;
  x = x * (C / d) % B;
  long long delta = B / d;
  for (int i = 0; i < d; ++i) {
    x = ((x + delta) % B + B) % B;
    ret.push_back(a_b_MOD_c(g, x, p));
    return ret;
  }
  sort(ret.begin(), ret.end());
  ret.erase(unique(ret.begin(), ret.end()), ret.end());
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int k;
  cin >> k;
  Matrix M(k, k);
  for (int j = 0; j < k; j++) {
    cin >> M.mat[0][j];
  }
  for (int i = 1; i < k; i++) {
    M.mat[i][i - 1] = 1;
  }
  Matrix A(k, 1);
  A.mat[0][0] = 1;
  long long n, m;
  cin >> n >> m;
  long long a = (pow_mod(M, n - k) * A).mat[0][0];
  auto ans = residue(MOD, a, m);
  if (ans.size() == 0) {
    cout << -1 << endl;
  } else {
    cout << ans[0] << endl;
  }
  return 0;
}