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 maxk = 101;
const int lbt = 998244353;
struct matrix {
  int a[maxk][maxk];
  int n;
  matrix() {
    n = 0;
    memset(a, 0, sizeof(a));
  }
  matrix(int n) {
    this->n = n;
    memset(a, 0, sizeof(a));
  }
  matrix operator*(const matrix &m) const {
    matrix ans;
    ans.n = n;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        for (int k = 1; k <= n; k++) {
          ans.a[i][k] =
              (ans.a[i][k] + (long long)a[i][j] * m.a[j][k]) % (lbt - 1);
        }
      }
    }
    return ans;
  }
};
matrix fast_pow(matrix m, int t);
int root(int t, int ans);
int fast_pow(int a, int t);
int rev(int a);
int b[maxk];
int k, n, m;
matrix g;
int main() {
  scanf("%d", &k);
  for (int i = 1; i <= k; i++) {
    scanf("%d", &b[i]);
  }
  scanf("%d%d", &n, &m);
  g.n = k;
  for (int i = 1; i < k; i++) {
    g.a[i + 1][i] = 1;
  }
  for (int i = 1; i <= k; i++) {
    g.a[k - i + 1][k] = b[i];
  }
  g = fast_pow(g, n - k);
  int t = g.a[k][k];
  printf("%d", root(t, m));
  return 0;
}
matrix fast_pow(matrix m, int t) {
  matrix ans;
  ans.n = m.n;
  for (int i = 1; i <= ans.n; i++) {
    ans.a[i][i] = 1;
  }
  for (int i = 31; i >= 0; i--) {
    ans = ans * ans;
    if (t >> i & 1) ans = ans * m;
  }
  return ans;
}
int fast_pow(int a, int t) {
  int ans = 1;
  for (int i = 31; i >= 0; i--) {
    ans = (long long)ans * ans % lbt;
    if (t >> i & 1) ans = (long long)ans * a % lbt;
  }
  return ans;
}
int bsgs(int base, int ans) {
  int sqtp = sqrt(lbt);
  int powrb = 1;
  int rb = rev(base);
  for (int i = 1; i <= sqtp; i++) {
    powrb = (long long)powrb * rb % lbt;
  }
  map<int, int> id;
  int powb = 1;
  for (int i = 0; i < sqtp; i++) {
    id[powb] = i;
    powb = (long long)powb * base % lbt;
  }
  for (int i = 0; i <= sqtp; i++) {
    if (id.count(ans)) return i * sqtp + id[ans];
    ans = (long long)ans * powrb % lbt;
  }
  return -1;
}
int d, x, y;
void gcd(int a, int b) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
    return;
  }
  gcd(b, a % b);
  int t = x;
  x = y;
  y = t - a / b * y;
}
int root(int t, int ans) {
  static int g = 3;
  int b = bsgs(g, ans);
  gcd(t, lbt - 1);
  if (b % d == 0) {
    if (x < 0) x += lbt - 1;
    int y = (long long)x * (b / d) % (lbt - 1);
    return fast_pow(g, y);
  } else {
    return -1;
  }
}
int rev(int a) {
  gcd(a, lbt);
  if (x < 0) x += lbt;
  return x;
}