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>
const int N = 105;
const int p = 998244353;
using namespace std;
int K;
int n, m;
void Add(int &x, int y, int p) {
  x += y;
  while (x >= p) x -= p;
}
struct data {
  int a[N][N], n, m;
  data operator*(const data &q) {
    data ans;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= q.m; j++) {
        ans.a[i][j] = 0;
        for (int k = 1; k <= m; k++)
          Add(ans.a[i][j], 1LL * a[i][k] * q.a[k][j] % (p - 1), p - 1);
      }
    ans.n = n;
    ans.m = q.m;
    return ans;
  }
} a, sum, ans;
void read(int &x) {
  char ch = getchar();
  x = 0;
  for (; ch < '0' || ch > '9'; ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 3) + (x << 1) + ch - '0';
}
void qpow(int n) {
  ans = a;
  n--;
  for (sum = a; n; n >>= 1, sum = sum * sum)
    if (n & 1) ans = ans * sum;
}
long long qpow(int a, int n) {
  long long ans = 1;
  for (long long sum = a; n; n >>= 1, sum = sum * sum % p)
    if (n & 1) ans = ans * sum % p;
  return ans;
}
void exgcd(int a, int b, int &g, int &x, int &y) {
  if (b == 0) {
    g = a;
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, g, y, x);
  y -= a / b * x;
}
namespace BSGS {
int hs[765433], head[765433], next[765433], id[765433], top;
void insert(int x, int y) {
  int k = x % 765433;
  hs[top] = x, id[top] = y, next[top] = head[k], head[k] = top++;
}
int find(int x) {
  int k = x % 765433;
  for (int i = head[k]; i != -1; i = next[i])
    if (hs[i] == x) return id[i];
  return -1;
}
int BSGS(int a, int b, int c) {
  memset(head, -1, sizeof(head));
  top = 1;
  if (b == 1) return 0;
  int m = sqrt(c * 1.0), j;
  long long x = 1, p = 1;
  for (int i = 0; i < m; ++i, p = p * a % c) insert(p * b % c, i);
  for (long long i = m;; i += m) {
    if ((j = find(x = x * p % c)) != -1) return i - j;
    if (i > c) break;
  }
  return -1;
}
}  // namespace BSGS
int main() {
  read(K);
  for (int i = 1; i <= K; i++) read(a.a[1][i]);
  for (int i = 2; i <= K; i++) a.a[i][i - 1] = 1;
  a.n = a.m = K;
  read(n);
  read(m);
  qpow(n - K);
  int v = 0;
  for (int i = 1; i <= 1; i++) Add(v, ans.a[1][i], p - 1);
  int w = BSGS::BSGS(3, m, p);
  int g, x, y;
  exgcd(v, p - 1, g, x, y);
  if (w % g)
    printf("-1\n");
  else {
    int a = 1ll * x * (w / g) % (p - 1);
    a = (a + p - 1) % (p - 1);
    cout << qpow(3, a) << endl;
  }
  return 0;
}