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 MAXK = 200;
const long long INF = 1e16 + 10;
const long long MOD = 998244353;
const long long g = 3;
int k, n, m, b[MAXK];
int fast_pow(int a, int p) {
  long long res = 1;
  for (long long t = a; p; t = (t * t) % MOD) {
    if (p & 1) res = (res * t) % MOD;
    p >>= 1;
  }
  return res;
}
int giant_small(long long a, long long b) {
  int sq = (int)sqrt(MOD) + 1;
  map<int, int> right;
  for (long long i = 0, ay = 1; i <= sq; ++i, ay = (ay * a) % MOD) {
    right[(ay * b) % MOD] = i;
  }
  for (int i = 0; i <= sq; ++i) {
    int ax = fast_pow(a, i * sq);
    if (right.count(ax)) {
      return i * sq - right[ax];
    }
  }
  return -1;
}
vector<vector<long long>> matr_mul(vector<vector<long long>> &a,
                                   vector<vector<long long>> &b,
                                   long long MOD) {
  vector<vector<long long>> res(k, vector<long long>(k));
  for (int i = 0; i < k; ++i) {
    for (int j = 0; j < k; ++j) {
      for (int z = 0; z < k; ++z) {
        res[i][j] = (res[i][j] + a[i][z] * b[z][j]) % MOD;
      }
    }
  }
  return res;
}
vector<vector<long long>> fast_pow(vector<vector<long long>> A, int p,
                                   long long MOD) {
  vector<vector<long long>> res(k, vector<long long>(k, 0));
  for (int i = 0; i < k; ++i) res[i][i] = 1;
  for (vector<vector<long long>> t = A; p; t = matr_mul(t, t, MOD)) {
    if (p & 1) res = matr_mul(res, t, MOD);
    p >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long gcd(long long a, long long &x, long long b, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  gcd(b, x, a % b, y);
  long long x1 = x;
  long long y1 = y;
  x = y1;
  y = x1 - a / b * y1;
}
int solve(long long a, long long b, long long c) {
  int gc = gcd(a, c);
  if (b % gc) {
    return -1;
  }
  a /= gc, b /= gc, c /= gc;
  long long x, y;
  gcd(a, x, c, y);
  if (x < 0) {
    x += (abs(x) + c - 1) / c * c;
  }
  return (x * b) % (MOD - 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> k;
  for (int i = 0; i < k; ++i) cin >> b[i];
  cin >> n >> m;
  int h_n = giant_small(g, m);
  vector<vector<long long>> A(k, vector<long long>(k));
  A[0] = vector<long long>(b, b + k);
  for (int i = 0; i < k - 1; ++i) A[i + 1][i] = 1;
  auto Ank = fast_pow(A, n - k, MOD - 1);
  int h_k = solve(Ank[0][0], h_n, MOD - 1);
  if (h_k < 0) {
    cout << -1;
  } else {
    cout << fast_pow(g, h_k);
  }
}