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 MOD1 = 998244353;
const int MOD2 = 998244352;
const int G = 3;
int read() {
  int v = 0, f = 1;
  char c = getchar();
  while (c < 48 || 57 < c) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (48 <= c && c <= 57) v = (v << 3) + v + v + c - 48, c = getchar();
  return v * f;
}
long long readll() {
  long long v = 0, f = 1;
  char c = getchar();
  while (c < 48 || 57 < c) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (48 <= c && c <= 57) v = (v << 3) + v + v + c - 48, c = getchar();
  return v * f;
}
int K;
struct Ma {
  long long a[111][111];
  void init() { memset(a, 0, sizeof(a)); }
} O;
void predo() {
  K = read();
  O.init();
  for (register int i = (int)1; i <= (int)K; ++i) O.a[i][i] = 1;
}
long long pw(long long x, long long y, long long p) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % p;
    x = x * x % p;
    y >>= 1;
  }
  return res;
}
long long bsgs(long long a, long long b, long long p) {
  long long m = ceil(sqrt(p));
  long long v = pw(a, m, p);
  v = pw(v, p - 2, p);
  long long i, e = 1;
  unordered_map<long long, long long> mp;
  mp[1] = 0;
  for (i = 1; i < m; i++) {
    e = e * a % p;
    if (!mp.count(e)) mp[e] = i;
  }
  for (i = 0; i < m; i++) {
    if (mp.count(b)) return i * m + mp[b];
    b = b * v % p;
  }
  return -2;
}
Ma mul(Ma a, Ma b) {
  Ma c;
  c.init();
  for (register int i = (int)1; i <= (int)K; ++i)
    for (register int j = (int)1; j <= (int)K; ++j)
      for (register int k = (int)1; k <= (int)K; ++k)
        c.a[i][j] = (c.a[i][j] + a.a[i][k] * b.a[k][j] % MOD2) % MOD2;
  return c;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b)
    x = 1, y = 0;
  else
    exgcd(b, a % b, y, x), y -= a / b * x;
}
long long inv(long long a, long long n) {
  long long x, y;
  exgcd(a, n, x, y);
  return (x + n) % n;
}
long long b[111], n, fn;
Ma A, B;
void init() {
  for (register int i = (int)1; i <= (int)K; ++i) b[i] = readll();
  n = read(), fn = read();
}
void solve() {
  A.init();
  A.a[1][K] = 1;
  B.init();
  for (register int i = (int)1; i <= (int)K; ++i)
    B.a[K + 1 - i][K] = b[i] % MOD2;
  for (register int i = (int)1; i <= (int)K - 1; ++i) B.a[i + 1][i] = 1;
  long long t = n - K;
  while (t) {
    if (t & 1) A = mul(A, B);
    B = mul(B, B);
    t >>= 1;
  }
  long long x = A.a[1][K];
  long long y = bsgs(G, fn, MOD1);
  if (y == -2) {
    printf("-1\n");
    return;
  }
  long long k = gcd(x, MOD2);
  if (y % k) {
    printf("-1\n");
    return;
  }
  x /= k;
  y /= k;
  long long tb = (y * inv(x, MOD2 / k) + MOD2 / k) % (MOD2 / k);
  long long fk = pw(G, tb, MOD1);
  printf("%lld\n", fk);
}
int main() {
  predo();
  int cas = 1;
  while (cas--) {
    init();
    solve();
  }
  return 0;
}