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 p = 998244353;
const int MAXN = 105;
namespace BSGS {
int hs[76543], head[76543], nxt[76543], id[76543], top;
void insert(int x, int y) {
  int k = x % 76543;
  hs[top] = x, id[top] = y, nxt[top] = head[k], head[k] = top++;
}
int find(int x) {
  int k = x % 76543;
  for (int i = head[k]; ~i; i = nxt[i]) {
    if (hs[i] == x) {
      return id[i];
    }
  }
  return -1;
}
int BSGS(int a, int b, int n) {
  memset(head, -1, sizeof(head));
  top = 1;
  if (b == 1) return 0;
  int m = sqrt(n * 1.0), j;
  long long x = 1, p = 1;
  for (int i = 0; i < m; ++i, p = p * a % n) insert(p * b % n, i);
  for (long long i = m;; i += m) {
    if ((j = find(x = x * p % n)) != -1) return i - j;
    if (i > n) break;
  }
  return -1;
}
long long main(long long a, long long b, long long n) {
  long long ans = BSGS(a, b, n);
  return ans;
}
}  // namespace BSGS
int K;
struct Matrix {
  int a[MAXN][MAXN];
  Matrix() {}
  Matrix(bool tag) {
    for (int i = 0; i < K; i++)
      for (int j = 0; j < K; j++) a[i][j] = 0;
    if (tag)
      for (int i = 0; i < K; i++) a[i][i] = 1;
  }
};
Matrix operator*(const Matrix &a, const Matrix &b) {
  Matrix ret(0);
  for (int k = 0; k < K; k++)
    for (int i = 0; i < K; i++)
      for (int j = 0; j < K; j++)
        ret.a[i][j] =
            (ret.a[i][j] + (long long)a.a[i][k] * b.a[k][j]) % (p - 1);
  return ret;
}
int b[MAXN], n, k;
long long a, bb, d;
int gcd(int x, int y) { return (y == 0 ? x : gcd(y, x % y)); }
int qpow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = (long long)ret * a % p;
    b >>= 1;
    a = (long long)a * a % p;
  }
  return ret;
}
long long ex_gcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return a;
  }
  long long q = ex_gcd(b, a % b, y, x);
  y -= a / b * x;
  return q;
}
int main() {
  scanf("%d", &K);
  for (int i = 0; i < K; i++) scanf("%d", &b[i]);
  scanf("%d%d", &k, &n);
  Matrix A(1), B(0), C(0);
  C.a[0][0] = 1;
  for (int j = 0; j < K; j++) B.a[0][j] = b[j];
  for (int j = 1; j < K; j++) B.a[j][j - 1] = 1;
  k -= K;
  while (k) {
    if (k & 1) A = A * B;
    k >>= 1;
    B = B * B;
  }
  C = A * C;
  k = C.a[0][0];
  a = BSGS::main(3, n, p);
  bb = p - 1;
  d = gcd(k, bb);
  if (a % d != 0)
    puts("-1");
  else {
    k /= d, bb /= d, a /= d;
    long long x, y;
    ex_gcd(k, bb, x, y);
    x = x * a % bb;
    if (x < 0) x += bb;
    printf("%d\n", qpow(3, x));
  }
  return 0;
}