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 zqm = 998244353, phi = zqm - 1, t = ceil(sqrt(zqm)), g = 3, K = 101;
int n;
struct mat {
  int d[K][K];
  inline const int* operator[](const int k) const { return d[k]; }
  inline int* operator[](const int k) { return d[k]; }
  inline mat(bool isunit = 0) {
    memset(d, 0, sizeof(d));
    for (int i = 1; i <= n; ++i) d[i][i] = isunit;
  }
} r;
inline mat operator*(const mat& lhs, const mat& rhs) {
  mat res;
  for (int k = 1; k <= n; ++k) {
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= n; ++j) {
        res[i][j] = (res[i][j] + lhs[i][k] * 1ll * rhs[k][j]) % phi;
      }
    }
  }
  return res;
}
inline mat qp(mat bas, int exp) {
  mat ret(1);
  for (; exp; exp /= 2, bas = bas * bas)
    if (exp & 1) ret = ret * bas;
  return ret;
}
inline int qp(int bas, int exp) {
  int ret = 1;
  for (; exp; exp /= 2, bas = (bas * 1ll * bas) % zqm)
    if (exp & 1) ret = (ret * 1ll * bas) % zqm;
  return ret;
}
inline int bsgs(int b) {
  static unordered_map<int, int> m;
  if (b == 1) return 0;
  int val = 1;
  m.clear();
  for (int v = 0; v < t; ++v, val = (val * 1ll * g) % zqm)
    m[(val * 1ll * b) % zqm] = v;
  int gt = val;
  for (int u = 1; u <= t; ++u, val = (val * 1ll * gt) % zqm)
    if (m.count(val)) return u * t - m[val];
  return -1;
}
int exgcd(int a, int b, int& x, int& y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  int d = exgcd(b, a % b, y, x);
  y -= (a / b) * x;
  return d;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &r[n - i + 1][n]);
  for (int i = 2; i <= n; ++i) r[i][i - 1] = 1;
  int m, x;
  scanf("%d%d", &m, &x);
  r = qp(r, m - n), x = bsgs(x);
  int inv, tmp, d = exgcd(r[n][n], phi, inv, tmp);
  if (x % d) {
    puts("-1");
    return 0;
  }
  cout << qp(g, ((x / d) * 1ll * inv % phi + phi) % phi) << '\n';
  return 0;
}