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;
template <typename T>
inline const T read1() {
  T m = 0;
  char k = getchar();
  while (('0' > k || k > '9') && (k != '-')) k = getchar();
  const bool f = (k == '-' ? 1 : 0);
  if (f) k = getchar();
  while ('0' <= k && k <= '9')
    m = (m << 3) + (m << 1) + (k ^ 48), k = getchar();
  return f ? -m : m;
}
class Mat {
  vector<vector<long long> > a;

 public:
  Mat() {}
  Mat(vector<vector<long long> > k) : a(k) {}
  Mat(long long x = 0, long long y = 0) {
    a.resize(x);
    for (long long i = 0; i < x; ++i) a[i].resize(y);
  }
  vector<long long>& operator[](const long long k) { return a[k]; }
  long long wide() { return a.size(); }
  long long len() { return a.empty() ? 0 : a[0].size(); }
  Mat operator*(Mat k) {
    vector<vector<long long> > tem;
    tem.resize(wide());
    for (long long i = 0; i < wide(); ++i) {
      tem[i].resize(k.len());
      for (long long j = 0; j < len(); ++j)
        for (long long l = 0; l < k.len(); ++l)
          (tem[i][l] += a[i][j] * k[j][l]) %= 998244353ll - 1;
    }
    return Mat(tem);
  }
  Mat operator+(Mat k) {
    long long o = max(wide(), k.wide()), p = max(len(), k.len());
    Mat tem(o, p);
    for (long long i = 0; i < o; ++i)
      for (long long j = 0; j < p; ++j) {
        if (i < wide() && j < len()) tem[i][j] = a[i][j];
        if (i < k.wide() && j < k.len()) tem[i][j] += k[i][j];
      }
    return tem;
  }
  Mat operator%(long long k) {
    Mat tem(wide(), len());
    for (long long i = 0; i < wide(); ++i)
      for (long long j = 0; j < len(); ++j) tem[i][j] = a[i][j] % k;
    return tem;
  }
  Mat& operator%=(long long k) { return *this = *this % k; }
  Mat& operator*=(Mat k) { return *this = *this * k; }
  Mat& operator+=(Mat k) { return *this = *this + k; }
  bool scan(long long x, long long y, const long long value) {
    if (x >= wide() || y >= len() || x < 0 || y < 0) return 0;
    a[x][y] = value;
    return 1;
  }
};
template <typename T>
T quickpow(T k, const long long n, bool ty = 1) {
  if (n == 1) return k;
  if (n == 0) return T(1);
  T tem = quickpow(k, n >> 1);
  tem = tem * tem;
  if (ty) tem %= 998244353ll;
  if (n & 1) {
    tem = tem * k;
    if (ty) tem %= 998244353ll;
  }
  return tem;
}
long long exgcd(long long a, long long b, long long& x, long long& y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  } else {
    long long t = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return t;
  }
}
map<long long, long long> ma;
long long s, k, g = 3, m;
long long bsgs(long long gt, long long x) {
  if (!(x % 998244353ll)) return -1;
  long long b = ceil(sqrt(998244353ll));
  long long t = 1, f = 1;
  for (long long i = 1; i <= b; ++i) {
    t = t * gt % 998244353ll;
    ma[t * x % 998244353ll] = i;
  }
  for (long long i = 1; i <= b; ++i)
    if (ma[f = f * t % 998244353ll]) return i * b - ma[f];
  return -1;
}
signed main() {
  k = read1<long long>();
  Mat a(k, k);
  for (long long i = 1; i < k; ++i) a[i][i - 1] = 1;
  for (long long i = k; i--;) a[i][k - 1] = read1<long long>();
  s = read1<long long>(), m = read1<long long>();
  a = quickpow(a, s - 1, 0);
  long long z = bsgs(quickpow(g, a[k - 1][0]), m);
  if (!~z) return puts("-1"), 0;
  printf("%lld", quickpow(g, z));
  return 0;
}