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 N = 2e5 + 10;
const long long M = 998244353;
struct Matrix {
  int n;
  long long P, a[205][205];
  Matrix() {}
  Matrix(int _n, long long _P) {
    n = _n;
    P = _P;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        a[i][j] = 0;
      }
    }
  }
  void init() {
    for (int i = 0; i < n; i++) {
      a[i][i] = 1;
    }
  }
  Matrix operator*(const Matrix &B) const {
    Matrix C(n, P);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < n; k++) {
          C.a[i][j] += a[i][k] * B.a[k][j] % P;
        }
        C.a[i][j] %= P;
      }
    }
    return C;
  }
  Matrix operator^(const long long &t) const {
    Matrix A = (*this), res = Matrix(n, P);
    res.init();
    long long tt = t;
    while (tt) {
      if (tt & 1) res = res * A;
      A = A * A;
      tt >>= 1;
    }
    return res;
  }
};
long long exgcd(long long a, long long b, long long &x, long long &y) {
  long long d = a;
  if (b != 0) {
    d = exgcd(b, a % b, y, x);
    y -= (a / b) * x;
  } else {
    x = 1;
    y = 0;
  }
  return d;
}
long long mod_inv(long long a, long long m) {
  long long x, y;
  exgcd(a, m, x, y);
  return (m + x % m) % m;
}
long long mod_pow(long long x, long long n, long long m) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % m;
    x = x * x % m;
    n >>= 1;
  }
  return res;
}
struct BSGS {
  long long a, t, P;
  map<long long, long long> mp;
  void init(long long _a, long long _P) {
    a = _a;
    P = _P;
    t = sqrt(P) + 1;
    long long at = mod_pow(a, t, P);
    mp.clear();
    long long x = 1, k = 0;
    while (k * t < P - 1) {
      mp[x] = k;
      k++;
      x = x * at % P;
    }
  }
  long long find(long long b) {
    long long inv_a = mod_pow(a, P - 2, P);
    for (long long B = 0, a_B = 1; B < t; B++, a_B = a_B * inv_a % P) {
      long long y = b * a_B % P;
      if (mp.find(y) != mp.end()) {
        return mp[y] * t + B;
      }
    }
    return -1;
  }
};
int k;
long long n, m, b[N];
long long solve(long long t, long long y) {
  long long g = 3;
  BSGS tool;
  tool.init(g, M);
  long long k = tool.find(y);
  long long a, b;
  long long d = exgcd(t, M - 1, a, b);
  if (k % d != 0) return -1;
  if (a < 0) {
    a = M - 1 + a;
    b = -b;
    b = t + b;
  }
  a = a * (k / d);
  b = b * (k / d);
  return mod_pow(g, a, M);
}
int main() {
  scanf("%d", &k);
  for (int i = 0; i < k; i++) {
    scanf("%lld", &b[i]);
  }
  scanf("%lld %lld", &n, &m);
  Matrix A(k, M - 1);
  for (int i = 0; i < k; i++) {
    if (i < k - 1) {
      A.a[i][i] = 0;
      A.a[i][i + 1] = 1;
    }
    A.a[k - 1][i] = b[k - 1 - i];
  }
  A = A ^ (n - k);
  long long t = A.a[k - 1][k - 1];
  printf("%lld\n", solve(t, m % M));
}