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 MOD = p - 1;
const int M = 105;
long long read() {
  long long x = 0, f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9') {
    if (c == '-') f = -1;
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
long long n, m, k, t, x, y;
struct Matrix {
  long long n, m, a[M][M];
  Matrix() {
    n = m = 0;
    memset(a, 0, sizeof a);
  }
  void clear() { memset(a, 0, sizeof a); }
  Matrix operator*(const Matrix &b) const {
    Matrix r;
    r.n = n;
    r.m = b.m;
    for (long long i = 1; i <= n; i++)
      for (long long j = 1; j <= m; j++)
        for (long long k = 1; k <= b.m; k++)
          r.a[i][k] = (r.a[i][k] + a[i][j] * b.a[j][k]) % MOD;
    return r;
  }
  void print() {
    for (long long i = 1; i <= n; i++, puts(""))
      for (long long j = 1; j <= m; j++) printf("%lld ", a[i][j]);
  }
} A;
Matrix qkpow(Matrix a, long long b) {
  Matrix r;
  r.n = r.m = k;
  for (long long i = 1; i <= k; i++) r.a[i][i] = 1;
  while (b > 0) {
    if (b & 1) r = r * a;
    a = a * a;
    b >>= 1;
  }
  return r;
}
long long bsgs(long long a, long long b) {
  long long m = sqrt(p) + 1, t = 1, tt = 1;
  map<long long, long long> mp;
  for (long long i = 0; i < m; i++) {
    mp[b * t % p] = i;
    t = t * a % p;
  }
  for (long long i = 1; i <= m; i++) {
    tt = tt * t % p;
    if (mp[tt]) return i * m - mp[tt];
  }
  return 0;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, y, x);
  y -= x * (a / b);
  return d;
}
long long fast(long long a, long long b) {
  long long r = 1;
  while (b > 0) {
    if (b & 1) r = r * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return r;
}
signed main() {
  k = read();
  A.n = A.m = k;
  for (long long i = 1; i <= k; i++) A.a[1][i] = read();
  for (long long i = 1; i <= k; i++) A.a[i + 1][i] = 1;
  n = read();
  m = read();
  A = qkpow(A, n - k);
  n = A.a[1][1];
  t = bsgs(3, m);
  long long d = exgcd(n, MOD, x, y);
  x = (x % MOD + MOD) % MOD;
  if (t % d)
    puts("-1");
  else {
    t /= d;
    printf("%lld\n", fast(3, t * x % MOD));
  }
}