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 k;
const int maxk = 105;
int b[maxk];
int n, m;
const int mod = 998244353;
const int mod1 = mod - 1;
void upd(int &a, int b, int M) {
  a += b;
  if (a >= M) a -= M;
}
struct mat {
  int a[maxk][maxk];
  mat() { memset(a, 0, sizeof(a)); }
  mat operator*(const mat b) const {
    mat c;
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        for (int l = 0; l < k; l++) {
          upd(c.a[i][l], (int)(1LL * a[i][j] * b.a[j][l] % mod1), mod1);
        }
      }
    }
    return c;
  }
};
mat power(mat a, int n) {
  mat I;
  for (int i = 0; i < k; i++) {
    I.a[i][i] = 1;
  }
  while (n > 0) {
    if (n & 1) I = (a * I);
    a = (a * a);
    n >>= 1;
  }
  return I;
}
int qpow(int a, int b, int mod) {
  int ret = 1;
  while (b > 0) {
    if (b & 1) ret = 1LL * ret * a % mod;
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return ret;
}
int bsgs(int a, int b, int p) {
  int m = ceil(sqrt(p));
  map<int, int> mp;
  int tmp = 1;
  mp[tmp] = 0;
  for (int i = 1; i < m; i++) {
    tmp = 1LL * tmp * a % p;
    mp[tmp] = i;
  }
  int ni = qpow(a, m, p);
  ni = qpow(ni, p - 2, p);
  int now = b;
  for (int i = 0; i <= m; i++) {
    if (mp.count(now)) {
      return (1LL * mp[now] + 1LL * m * i) % (p - 1);
    }
    now = 1LL * now * ni % p;
  }
  return -1;
}
void gcd(long long a, long long b, long long &d, long long &x, long long &y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    gcd(b, a % b, d, y, x);
    y -= 1LL * x * (a / b);
  }
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  while (cin >> k) {
    for (int i = 0; i < k; i++) {
      cin >> b[i];
    }
    cin >> n >> m;
    mat a;
    for (int i = 1; i < k; i++) {
      a.a[i][i - 1] = 1;
    }
    for (int j = 0; j < k; j++) {
      a.a[0][j] = b[j];
    }
    a = power(a, n - k);
    int pw = a.a[0][0];
    int pw1 = bsgs(3, m, mod);
    long long g = gcd(1LL * pw, 1LL * mod1);
    if (pw1 % g)
      cout << -1 << endl;
    else {
      long long d, x, y;
      long long A, B;
      A = pw / g;
      B = mod1 / g;
      pw1 /= g;
      gcd(A, B, d, x, y);
      x *= pw1;
      y *= pw1;
      x = ((x % B) + B) % B;
      cout << qpow(3, (int)x, mod) << endl;
    }
  }
  return 0;
}