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 MAXN = 105;
const int p = 998244353;
const int g = 3;
struct Matrix {
  int r, c;
  int a[MAXN][MAXN];
  Matrix(int R = 0, int C = 0) {
    r = R;
    c = C;
    memset(a, 0, sizeof(a));
  }
  Matrix operator*(const Matrix &t) const {
    Matrix res(r, t.c);
    for (int k = 1; k <= c; k++)
      for (int i = 1; i <= res.r; i++) {
        long long r = a[i][k];
        for (int j = 1; j <= res.c; j++)
          res.a[i][j] = (r * t.a[k][j] + res.a[i][j]) % (p - 1);
      }
    return res;
  }
} trans, st;
int k, n, a, b;
map<int, int> hs;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
void exgcd(int a, int b, int &x, int &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, y, x);
  y -= a / b * x;
}
int power(int a, int b, int mod) {
  int res = 1;
  while (b) {
    if (b & 1) res = (long long)res * a % mod;
    a = (long long)a * a % mod;
    b >>= 1;
  }
  return res;
}
Matrix power(Matrix a, int b) {
  Matrix res = a;
  b--;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int inverse(int a, int mod) {
  int x, y;
  exgcd(a, mod, x, y);
  return (x % mod + mod) % mod;
}
int BSGS(int g, int b, int mod) {
  int sz = (int)ceil(sqrt(mod) + 1), cur = 1;
  for (int i = 0; i < sz; i++) {
    if (cur == b) return i;
    hs[cur] = i + 1;
    cur = (long long)cur * g % mod;
  }
  int base = inverse(cur, mod);
  cur = (long long)base * b % mod;
  for (int i = sz; i <= mod - 1; i += sz) {
    int j = hs[cur] - 1;
    if (j != -1) return i + j;
    cur = (long long)cur * base % mod;
  }
  return -1;
}
int main() {
  cin >> k;
  trans = Matrix(k, k);
  st = Matrix(1, k);
  for (int i = 1; i <= k; i++) cin >> trans.a[i][1];
  for (int i = 2; i <= k; i++) trans.a[i - 1][i] = 1;
  st.a[1][1] = 1;
  cin >> n >> b;
  st = st * power(trans, n - k);
  a = st.a[1][1];
  int indb = BSGS(g, b, p), mod = p - 1, d = gcd(a, mod);
  if (indb % d != 0) {
    cout << -1 << endl;
    return 0;
  }
  cout << power(g, (long long)indb / d * inverse(a / d, mod / d) % (mod / d), p)
       << endl;
  return 0;
}