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 long long N = 1e6 + 10;
const long long M = 1e2 + 10;
const long long mod = 998244353;
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 gd = exgcd(b, a % b, y, x);
  y -= a / b * x;
  return gd;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long inv(long long a, long long b) {
  long long x, y;
  exgcd(a, b, x, y);
  return (x % b + b) % b;
}
long long qpow(long long a, long long x, long long mo) {
  long long res = 1;
  while (x) {
    if (x & 1) res = 1ll * res * a % mo;
    a = 1ll * a * a % mo;
    x >>= 1;
  }
  return res;
}
struct Matrix {
  long long a[M][M];
  long long n, m;
} e;
Matrix mat_mul(Matrix x, Matrix y, long long P) {
  Matrix res;
  memset(res.a, 0, sizeof(res.a));
  res.n = x.n, res.m = y.m;
  for (long long i = 1; i <= res.n; i++)
    for (long long j = 1; j <= res.m; j++)
      for (long long k = 1; k <= x.m; k++)
        res.a[i][j] = (res.a[i][j] + 1ll * x.a[i][k] * y.a[k][j] % P) % P;
  return res;
}
Matrix mqpow(Matrix x, long long y, long long P) {
  Matrix ans = e;
  while (y) {
    if (y & 1) ans = mat_mul(ans, x, P);
    x = mat_mul(x, x, P);
    y >>= 1;
  }
  return ans;
}
long long BSGS(long long a, long long b, long long p) {
  map<long long, long long> mp;
  long long k = sqrt(p) + 1;
  long long tmp = b;
  long long pa = 1;
  for (long long i = 1; i <= k; i++) {
    tmp = 1ll * tmp * a % p;
    pa = 1ll * pa * a % p;
    mp[tmp] = i;
  }
  long long now = 1;
  for (long long i = 1; i <= k; i++) {
    now = 1ll * now * pa % p;
    if (mp[now]) {
      return (i * k - mp[now] + 2 * p) % p;
    }
  }
  return -1;
}
long long b[N];
int main() {
  long long k, n, m;
  cin >> k;
  for (long long i = 1; i <= k; i++) cin >> b[i];
  for (long long i = 1; i <= k; i++) e.a[i][i] = 1;
  cin >> n >> m;
  Matrix xx;
  xx.n = xx.m = e.n = e.m = k;
  for (long long i = 1; i <= k; i++) xx.a[1][i] = b[i] % (mod - 1);
  for (long long i = 1; i < k; i++) xx.a[i + 1][i] = 1;
  Matrix yy;
  yy.n = k, yy.m = 1;
  for (long long i = 1; i <= k; i++) yy.a[i][1] = 0;
  yy.a[1][1] = 1;
  long long res = mat_mul(mqpow(xx, n - k, mod - 1), yy, mod - 1).a[1][1];
  {
    long long ans = BSGS(3, m, mod);
    if (ans == -1)
      puts("-1");
    else {
      if (ans % gcd(mod - 1, res)) {
        puts("-1");
      } else {
        long long x, y;
        exgcd(res, mod - 1, x, y);
        x = ((x % (mod - 1) + mod - 1) % (mod - 1)) *
            (ans / gcd(mod - 1, res)) % (mod - 1);
        cout << qpow(3, x, mod) << endl;
      }
    }
  }
}