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;
int c[10005] = {0}, bb[10005] = {0}, tmp[20005] = {0};
int k, n, m, b[105];
const int p = 998244353;
void mul(int *a1, int *a2) {
  for (int i = 0; i <= (k << 1); ++i) tmp[i] = 0;
  for (int i = 0; i <= k; ++i) {
    for (int j = 0; j <= k; ++j) {
      tmp[i + j] = (tmp[i + j] + 1ll * a1[i] * a2[j]) % (p - 1);
    }
  }
  for (int i = k << 1; i >= k; --i) {
    for (int j = 0; j < k; ++j) {
      tmp[i - k + j] = (tmp[i - k + j] + 1ll * tmp[i] * b[k - j]) % (p - 1);
    }
    tmp[i] = 0;
  }
  for (int i = 0; i < k; ++i) a1[i] = tmp[i];
}
void poww(int *a1, int t, int *a2) {
  while (t) {
    if (t & 1) mul(a2, a1);
    mul(a1, a1), t >>= 1;
  }
}
int poww_i(int a, int b, int c) {
  a %= c;
  int res = 1;
  while (b) {
    if (b & 1) res = (1ll * res * a) % c;
    a = (1ll * a * a) % c, b >>= 1;
  }
  return res;
}
int extgcd(int a, int b, int &x, int &y) {
  int d = a;
  if (b)
    d = extgcd(b, a % b, y, x), y = y - (a / b) * x;
  else
    x = 1, y = 0;
  return d;
}
int inv(int x, int m) {
  int u, v;
  if (extgcd(x, m, u, v) != 1) return -1;
  return u < 0 ? u + m : u;
}
int BSGS(int y, int z, int m) {
  map<int, int> rec;
  int s = (int)floor(sqrt(m - 1) + 0.5), cur = 1;
  int v = inv(poww_i(y, s, m), m);
  if (v == -1) return -1;
  for (int i = 0; i < s; ++i) rec[cur] = i, cur = 1ll * cur * y % m;
  for (int i = 0; i <= s; ++i) {
    if (rec.count(z)) return i * s + rec[z];
    z = 1ll * z * v % m;
  }
  return -1;
}
int main() {
  scanf("%d", &k);
  for (int i = 1; i <= k; ++i) scanf("%d", &b[i]);
  scanf("%d%d", &n, &m);
  bb[1] = 1, c[0] = 1;
  if (k > 1)
    poww(bb, n - 1, c);
  else
    c[k - 1] = poww_i(b[1], n - 1, p - 1);
  int exp1 = BSGS(3, m, p), u, v;
  int d = extgcd(c[k - 1], p - 1, u, v);
  if (exp1 % d)
    printf("-1\n");
  else {
    u = (u + p - 1) % (p - 1);
    u = 1ll * (exp1 / d) * u % (p - 1);
    printf("%d\n", poww_i(3, u, p));
  }
  return 0;
}