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 long long int p = 998244353;
long long int extgcd(long long int a, long long int b, long long int& x,
                     long long int& y) {
  long long int d = a;
  if (b != 0) {
    d = extgcd(b, a % b, y, x);
    y -= (a / b) * x;
  } else {
    x = 1, y = 0;
  }
  return d;
}
long long int inv(long long int a, long long int p) {
  long long int x, y;
  extgcd(a, p, x, y);
  if (x < 0) {
    return p - (-x % p);
  } else {
    return x % p;
  }
}
long long int powmod(long long int a, long long int k, long long int mod) {
  long long int ap = a, ans = 1;
  while (k) {
    if (k & 1) {
      ans *= ap;
      ans %= mod;
    }
    ap = ap * ap;
    ap %= mod;
    k >>= 1;
  }
  return ans;
}
long long int discretelog(long long int g, long long int y, long long int p) {
  if (g % p == 0) {
    if (y % p == 0) {
      return 0;
    } else {
      return -1;
    }
  }
  if (p == 2) {
    if (y % 2 == 0) {
      return -1;
    } else {
      return 0;
    }
  }
  long long int m = (long long int)(sqrt((double)p) + 1.0);
  g %= p;
  map<long long int, long long int> gpow;
  long long int gp = 1;
  gpow[1] = 0;
  for (long long int i = 1; i < m; i++) {
    gp = gp * g % p;
    gpow[gp] = i;
  }
  long long int gm = powmod(g, p - 1 - m, p);
  long long int r = y % p;
  for (long long int i = 0; i < m; i++) {
    auto itr = gpow.find(r);
    if (itr != gpow.end()) {
      return (i * m + (itr->second)) % (p - 1);
    }
    r = r * gm % p;
  }
  return -1;
}
vector<vector<long long int>> matrixmul(int l, int m, int n,
                                        vector<vector<long long int>> a,
                                        vector<vector<long long int>> b) {
  vector<vector<long long int>> c;
  for (int i = 0; i < l; i++) {
    vector<long long int> v;
    for (int j = 0; j < n; j++) {
      long long int x = 0;
      for (int k = 0; k < m; k++) {
        x += (a[i][k] * b[k][j]);
        x %= (p - 1);
      }
      v.push_back(x);
    }
    c.push_back(v);
  }
  return c;
}
vector<vector<long long int>> matrixpow(int n, vector<vector<long long int>> a,
                                        long long int k) {
  vector<vector<long long int>> ap = a, ans;
  for (int i = 0; i < n; i++) {
    vector<long long int> v;
    for (int j = 0; j < n; j++) {
      if (i == j) {
        v.push_back(1);
      } else {
        v.push_back(0);
      }
    }
    ans.push_back(v);
  }
  while (k) {
    if (k & 1) ans = matrixmul(n, n, n, ap, ans);
    ap = matrixmul(n, n, n, ap, ap);
    k >>= 1;
  }
  return ans;
}
int main() {
  int k;
  cin >> k;
  vector<long long int> b;
  for (int i = 0; i < k; i++) {
    long long int b1;
    cin >> b1;
    b.push_back(b1);
  }
  long long int n, m;
  cin >> n >> m;
  vector<vector<long long int>> a;
  a.push_back(b);
  for (int i = 1; i < k; i++) {
    vector<long long int> a1;
    for (int j = 0; j < k; j++) {
      if (j == i - 1)
        a1.push_back(1);
      else
        a1.push_back(0);
    }
    a.push_back(a1);
  }
  vector<vector<long long int>> ap = matrixpow(k, a, n - (long long int)k);
  long long int lm = discretelog(3, m, p);
  long long int la = ap[0][0];
  long long int x, y;
  long long int d = extgcd(la, p - 1, x, y);
  if (lm % d != 0) {
    cout << -1 << endl;
    return 0;
  }
  x = (x * (lm / d) % (p - 1) + (p - 1)) % (p - 1);
  cout << powmod(3, x, p) << endl;
  return 0;
}