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 K = 110;
const int MOD = 998244353;
const int PHI = MOD - 1;
int _w;
void add_phi(int &a, int b) {
  a += b;
  a -= a >= PHI ? PHI : 0;
}
int pow_mod(int a, int b, int p) {
  if (!b) return 1;
  long long rtn = pow_mod(a, b >> 1, p);
  rtn = rtn * rtn % p;
  if (b & 1) rtn = rtn * a % p;
  return int(rtn);
}
int inv(int x, int p) { return pow_mod(x, p - 2, p); }
void exgcd(int a, int b, int &d, long long &x, long long &y) {
  if (!b) {
    d = a, x = 1, y = 0;
  } else {
    exgcd(b, a % b, d, y, x);
    y -= a / b * x;
  }
}
int k, b[K], n, m;
int A[K][K], Q[K][K];
void mul(int A[K][K], int B[K][K], int C[K][K]) {
  static int tmp[K][K];
  memset(tmp, 0, sizeof tmp);
  for (int i = 0; i < k; ++i)
    for (int j = 0; j < k; ++j)
      for (int kk = 0; kk < k; ++kk)
        add_phi(tmp[i][j], int(1LL * A[i][kk] * B[kk][j] % PHI));
  for (int i = 0; i < k; ++i)
    for (int j = 0; j < k; ++j) C[i][j] = tmp[i][j];
}
void fpow(int A[K][K], int b, int C[K][K]) {
  static int tmp[K][K];
  memset(tmp, 0, sizeof tmp);
  for (int i = 0; i < k; ++i) tmp[i][i] = 1;
  while (b) {
    if (b & 1) mul(tmp, A, tmp);
    mul(A, A, A);
    b >>= 1;
  }
  for (int i = 0; i < k; ++i)
    for (int j = 0; j < k; ++j) C[i][j] = tmp[i][j];
}
int solve() {
  A[0][k - 1] = 1;
  for (int i = 1; i < k; ++i) Q[i][i - 1] = 1;
  for (int i = 0; i < k; ++i) Q[i][k - 1] = b[i] % PHI;
  fpow(Q, n - k, Q);
  mul(A, Q, A);
  return A[0][k - 1];
}
map<long long, int> mp;
int bsgs(int a, int b, int p) {
  mp.clear();
  int m = int(sqrt(p) + 1);
  long long e = 1;
  for (int i = 0; i < m; ++i) {
    if (!mp.count(e)) mp[e] = i;
    e = e * a % p;
  }
  long long iv = inv(int(e), p);
  int orb = b;
  for (int i = 0; i < m; ++i) {
    if (mp.count(b)) {
      int x = i * m + mp[b];
      if (pow_mod(a, x, p) == orb) return x;
    }
    b = int(b * iv % p);
  }
  return -1;
}
void kleft(int P, int A, int B) {
  int g = 3, bk = bsgs(g, B, P), phip = P - 1;
  int d;
  long long xk, n;
  exgcd(A, -phip, d, xk, n);
  if (bk % d) {
    puts("-1");
    return;
  }
  xk *= bk / d, n *= bk / d;
  int phip2 = abs(phip / d);
  xk = (xk % phip2 + phip2) % phip2;
  mp.clear();
  long long x = pow_mod(g, int(xk), P);
  printf("%lld\n", x);
}
int main() {
  _w = scanf("%d", &k);
  for (int i = 0; i < k; ++i) _w = scanf("%d", b + i);
  reverse(b, b + k);
  _w = scanf("%d%d", &n, &m);
  int tmp = solve();
  kleft(MOD, tmp, m);
  return 0;
}