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>
int a;
std::vector<std::vector<long long int> > V2;
void power(std::vector<std::vector<long long int> > &V, int t) {
  if (t == 1) return;
  if (t % 2 == 0) {
    power(V, t / 2);
    for (int i = 1; i <= a; i++)
      for (int j = 1; j <= a; j++) V2[i][j] = 0;
    for (int i = 1; i <= a; i++) {
      for (int j = 1; j <= a; j++) {
        for (int k = 1; k <= a; k++) {
          V2[i][j] += V[i][k] * V[k][j];
          V2[i][j] %= (998244353 - 1);
        }
      }
    }
    for (int i = 1; i <= a; i++)
      for (int j = 1; j <= a; j++) V[i][j] = V2[i][j];
  } else {
    std::vector<std::vector<long long int> > temp;
    temp.resize(a + 1);
    for (int i = 0; i <= a; i++) temp[i].resize(a + 1);
    for (int i = 1; i <= a; i++)
      for (int j = 1; j <= a; j++) temp[i][j] = V[i][j];
    power(V, t / 2);
    for (int i = 1; i <= a; i++)
      for (int j = 1; j <= a; j++) V2[i][j] = 0;
    for (int i = 1; i <= a; i++) {
      for (int j = 1; j <= a; j++) {
        for (int k = 1; k <= a; k++) {
          V2[i][j] += V[i][k] * V[k][j];
          V2[i][j] %= (998244353 - 1);
        }
      }
    }
    for (int i = 1; i <= a; i++)
      for (int j = 1; j <= a; j++) V[i][j] = 0;
    for (int i = 1; i <= a; i++) {
      for (int j = 1; j <= a; j++) {
        for (int k = 1; k <= a; k++) {
          V[i][j] += V2[i][k] * temp[k][j];
          V[i][j] %= (998244353 - 1);
        }
      }
    }
  }
}
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  long long int k = a;
  while (b) {
    if (b % 2 == 1) ans *= k, ans %= 998244353;
    k *= k, k %= 998244353;
    b /= 2;
  }
  return ans;
}
std::map<long long int, long long int> M, M2;
std::vector<std::vector<long long int> > V;
int main() {
  scanf("%d", &a);
  V.resize(a + 1);
  for (int i = 0; i <= a; i++) V[i].resize(a + 1);
  V2.resize(a + 1);
  for (int i = 0; i <= a; i++) V2[i].resize(a + 1);
  for (int i = 1; i <= a; i++) {
    int b;
    scanf("%d", &b);
    V[a][a - i + 1] = b;
  }
  for (int i = 1; i < a; i++) V[i][i + 1] = 1;
  int b, c;
  scanf("%d%d", &b, &c);
  power(V, b - 1);
  long long int d = -1, ans = -1;
  for (int i = 1; i < 40000; i++) {
    long long int t = power(3, 40000 * i);
    M[t] = i;
  }
  for (int i = 0; i < 40000; i++) {
    long long int t = power(3, i);
    t *= c, t %= 998244353;
    if (M.find(t) != M.end()) {
      d = 40000 * M[t] - i;
      d %= (998244353 - 1);
      break;
    }
  }
  for (int i = 1; i < 40000; i++) {
    long long int t = (40000 * V[1][a] % (998244353 - 1)) * i;
    t %= (998244353 - 1);
    M2[t] = i;
  }
  for (int i = 0; i < 40000; i++) {
    long long int t = V[1][a] * i + d;
    t %= (998244353 - 1);
    if (M2.find(t) != M2.end()) {
      ans = 40000 * M2[t] - i;
      ans %= (998244353 - 1);
      break;
    }
  }
  if (ans != -1)
    printf("%lld", power(3, ans));
  else
    printf("-1");
}