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 double pi = acos(-1.0);
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const int maxn = 105;
int mod = 998244353;
int g = 3, k, n, m;
map<int, int> ma;
int b[maxn];
struct matrix {
  int c[maxn][maxn];
  int n;
  matrix() {
    this->n = k;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        c[i][j] = 0;
      }
    }
  }
  matrix operator*(const matrix &m) const {
    matrix res;
    for (int i = 1; i <= n; i++) {
      for (int k = 1; k <= n; k++) {
        for (int j = 1; j <= n; j++) {
          res.c[i][j] += 1ll * c[i][k] * m.c[k][j] % mod;
          res.c[i][j] %= mod;
        }
      }
    }
    return res;
  }
};
matrix qpow(matrix a, long long p) {
  matrix res;
  for (int i = 1; i <= k; i++) res.c[i][i] = 1;
  while (p) {
    if (p & 1) res = res * a;
    a = a * a;
    p >>= 1;
  }
  return res;
}
long long qpow(long long a, long long p, long long mod) {
  long long res = 1;
  while (p) {
    if (p & 1) res = res * a % mod;
    a = a * a % mod;
    p >>= 1;
  }
  return res;
}
int bsgs(int y, int z, int p) {
  ma.clear();
  int ans = -1, m = (int)ceil(sqrt(p)), s = z;
  for (int i = 0; i < m; i++) {
    ma[s] = i;
    s = 1ll * s * y % p;
  }
  int buf = qpow(y, m, p);
  s = 1;
  for (int i = 1; i <= m; i++) {
    s = 1ll * s * buf % p;
    if (ma.count(s)) {
      ans = i * m - ma[s];
      break;
    }
  }
  return ans;
}
void exgcd(long long a, long long b, long long &d, long long &x, long long &y) {
  if (b == 0) {
    d = a;
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, d, y, x);
  y -= (a / b) * x;
}
int main(void) {
  scanf("%d", &k);
  for (int i = 1; i <= k; i++) {
    scanf("%d", &b[i]);
  }
  scanf("%d%d", &n, &m);
  int hn = bsgs(g, m, mod);
  matrix mat;
  for (int i = 1; i <= k; i++) {
    mat.c[1][i] = b[i];
  }
  for (int i = 2; i <= k; i++) {
    mat.c[i][i - 1] = 1;
  }
  --mod;
  matrix res = qpow(mat, n - k);
  int val = res.c[1][1];
  long long d, x, y;
  exgcd(val, mod, d, x, y);
  if (hn % d != 0) {
    puts("-1");
    return 0;
  }
  long long hk = 1ll * x * (hn / d) % (mod / d);
  hk = (hk % (mod / d) + (mod / d)) % (mod / d);
  ++mod;
  int fk = qpow(g, hk, mod);
  printf("%d\n", fk);
  return 0;
}