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 mod = 998244353, Mo = 100003;
int n, m, k;
struct Matrix {
  int a[103][103];
  Matrix() { memset(a, 0, sizeof(a)); }
  int *operator[](int x) { return a[x]; }
  Matrix operator*(Matrix A) {
    Matrix C;
    for (int w = 1; w <= k; w++)
      for (int i = 1; i <= k; i++)
        for (int j = 1; j <= k; j++)
          (C[i][j] += 1ll * a[i][w] * A[w][j] % (mod - 1)) %= (mod - 1);
    return C;
  }
} Ans, Base;
int ksm(int x, int k) {
  int s = 1;
  for (; k; k >>= 1, x = 1ll * x * x % mod)
    if (k & 1) s = 1ll * s * x % mod;
  return s;
}
int gcd(int a, int b) { return !b ? a : gcd(b, a % b); }
void exgcd(int a, int b, int &x, int &y) {
  if (!b)
    x = 1, y = 0;
  else
    exgcd(b, a % b, y, x), y -= a / b * x;
}
struct Hstb {
  vector<pair<int, int> > H[Mo];
  void ins(int x, int i) { H[x % Mo].push_back(make_pair(x, i)); }
  int find(int x) {
    for (auto w : H[x % Mo])
      if (w.first == x) return w.second;
    return -1;
  }
} Hs;
int BSGS(int g, int A, int P) {
  int M = sqrt(P);
  for (int i = 0, tt = A; i <= M; i++, tt = 1ll * tt * g % P) Hs.ins(tt, i);
  for (int i = 1, bs = ksm(g, M), tt = bs; i <= M + 1;
       i++, tt = 1ll * tt * bs % P)
    if (Hs.find(tt) != -1) return i * M - Hs.find(tt);
  return -1;
}
signed main() {
  cin >> k;
  for (int i = 1; i <= k; i++) cin >> Base[k - i + 1][k];
  for (int i = 2; i <= k; i++) Base[i][i - 1] = 1;
  Ans[1][k] = 1;
  cin >> n >> m;
  n--;
  for (; n; n >>= 1, Base = Base * Base)
    if (n & 1) Ans = Ans * Base;
  int r = BSGS(3, m, mod);
  int a = Ans[1][1], b = mod - 1, g = gcd(a, b), x, y;
  if (r % g) return puts("-1"), 0;
  a /= g, b /= g;
  r /= g;
  exgcd(a, b, x, y);
  x = 1ll * x * r % (mod - 1);
  int c = (x + mod - 1) % (mod - 1);
  x = ksm(3, c);
  cout << x << endl;
}