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 mod = 998244353;
const int g = 3;
struct Mat {
  int n, m;
  int a[105][105];
  Mat() { memset(a, 0, sizeof(a)); }
  Mat operator*(const Mat &oth) const {
    Mat ans;
    ans.n = n;
    ans.m = oth.m;
    for (int i = 0; i < (int)(ans.n); ++i)
      for (int j = 0; j < (int)(ans.m); ++j)
        for (int k = 0; k < (int)(m); ++k)
          (ans.a[i][j] += (long long)a[i][k] * oth.a[k][j] % (mod - 1)) %=
              mod - 1;
    return ans;
  }
};
Mat pow(Mat x, int n) {
  Mat ans;
  ans.n = x.n;
  ans.m = x.m;
  for (int i = 0; i < (int)(ans.n); ++i) ans.a[i][i] = 1;
  for (; n; n >>= 1, x = x * x)
    if (n & 1) ans = ans * x;
  return ans;
}
int pow(int x, int n) {
  int ans = 1;
  for (; n; n >>= 1, x = (long long)x * x % mod)
    if (n & 1) ans = (long long)ans * x % mod;
  return ans;
}
const int M = 32768;
unordered_map<int, int> REM;
void init_dlog() {
  int x = 1;
  for (int i = 0; i < (int)(M); ++i) x = (long long)x * g % mod;
  int y = 1;
  for (int i = 0; i < (int)(M); ++i) {
    if (!REM.count(y)) REM[y] = i * M;
    y = (long long)y * x % mod;
  }
}
int dlog(int x) {
  int y = x;
  for (int i = 0; i < (int)(M); ++i) {
    if (REM.count(y)) return ((REM[y] - i) % (mod - 1) + (mod - 1)) % (mod - 1);
    y = (long long)y * g % mod;
  }
  return -1;
}
int gcd(int a, int b) { return !b ? a : gcd(b, a % b); }
int exgcd(int a, int b, int &x, int &y) {
  if (!b) {
    x = 1;
    y = 0;
    return a;
  } else {
    int d = exgcd(b, a % b, y, x);
    y -= (a / b) * x;
    return d;
  }
}
int inv(int x, int mod) {
  int a, b;
  int d = exgcd(x, mod, a, b);
  assert(d == 1);
  return (a % mod + mod) % mod;
}
int k;
int b[105];
int n, m;
int main() {
  scanf("%d", &k);
  for (int i = 0; i < (int)(k); ++i) scanf("%d", &b[i]), b[i] %= mod - 1;
  scanf("%d %d", &n, &m);
  init_dlog();
  int logm = dlog(m);
  Mat trans;
  trans.n = k;
  trans.m = k;
  for (int i = 0; i < (int)(k); ++i) trans.a[0][i] = b[i];
  for (int i = 0; i < (int)(k - 1); ++i) trans.a[i + 1][i] = 1;
  trans = pow(trans, n - k);
  Mat ini;
  ini.n = k;
  ini.m = 1;
  ini.a[0][0] = 1;
  trans = trans * ini;
  int c = trans.a[0][0];
  int gc = gcd(c, mod - 1);
  if (logm % gc != 0) {
    puts("-1");
    return 0;
  }
  int x = (long long)inv(c / gc, (mod - 1) / gc) * (logm / gc) % (mod - 1);
  printf("%d\n", pow(g, x));
  return 0;
}