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;
class Solution {
 public:
  int solve(vector<int>& coes, int k, int n, int m) {
    vector<vector<int>> mat(k, vector<int>(k, 0));
    for (int i = 0; i < k; ++i) {
      mat[0][i] = coes[i] % (M - 1);
    }
    for (int i = 1; i < k; ++i) {
      mat[i][i - 1] = 1;
    }
    n -= k;
    vector<vector<int>> res(k, vector<int>(k, 0));
    for (int i = 0; i < k; ++i) {
      res[i][i] = 1;
    }
    while (n > 0) {
      if (n & 0x1) {
        res = matrixMul(res, mat, M - 1);
      }
      mat = matrixMul(mat, mat, M - 1);
      n >>= 1;
    }
    int s = res[0][0];
    int t = search(P, m, M);
    int ans = calc(s, t, M - 1);
    if (ans < 0) {
      return -1;
    } else {
      return powM(P, ans, M);
    }
  }

 private:
  const int M = 998244353;
  const int P = 3;
  vector<vector<int>> matrixMul(vector<vector<int>>& a, vector<vector<int>>& b,
                                int M) {
    int n = a.size();
    assert(n != 0);
    int m = a[0].size();
    assert(m != 0);
    assert(m == b.size());
    int l = b[0].size();
    vector<vector<int>> res(n, vector<int>(l, 0));
    for (int i = 0; i < n; ++i) {
      for (int k = 0; k < m; ++k) {
        for (int j = 0; j < l; ++j) {
          res[i][j] += (a[i][k] * 1LL * b[k][j]) % M;
          if (res[i][j] >= M) {
            res[i][j] -= M;
          }
        }
      }
    }
    return res;
  }
  long long normalize(long long a, long long M) {
    a %= M;
    if (a < 0) {
      a += M;
    }
    return a;
  }
  long long mulM(long long a, long long b, long long M) {
    return normalize(a * b, M);
  }
  long long powM(long long b, long long e, long long M) {
    long long res = 1;
    while (e > 0) {
      if (e & 0x1) {
        res = mulM(res, b, M);
      }
      b = mulM(b, b, M);
      e >>= 1;
    }
    return res;
  }
  long long invM(long long a, long long M) {
    a = normalize(a, M);
    long long b = M;
    long long u = 0, v = 1;
    while (a != 0) {
      long long d = b / a;
      b -= d * a;
      u -= d * v;
      swap(a, b);
      swap(u, v);
    }
    assert(b == 1);
    return normalize(u, M);
  }
  int search(int a, int b, int M) {
    map<int, int> mp;
    int m = sqrt(M);
    while (m * 1LL * m < M) {
      ++m;
    }
    long long y = 1;
    for (int i = 0; i < m; ++i) {
      int key = mulM(b, invM(y, M), M);
      if (mp.count(key) == 0) {
        mp[key] = i;
      }
      y = mulM(y, a, M);
    }
    long long x = 1;
    for (int i = 0; i < m; ++i) {
      if (mp.count(x) > 0) {
        return i * m + mp[x];
      }
      x = mulM(x, y, M);
    }
    assert(0);
    return -1;
  }
  long long egcd(long long a, long long b, long long& x, long long& y) {
    int ax = 1;
    int ay = 0;
    int bx = 0;
    int by = 1;
    while (a != 0) {
      long long d = b / a;
      b -= d * a;
      bx -= d * ax;
      by -= d * ay;
      swap(a, b);
      swap(ax, bx);
      swap(ay, by);
    }
    x = bx;
    y = by;
    return b;
  }
  long long gcd(long long a, long long b) {
    while (a != 0) {
      b %= a;
      swap(a, b);
    }
    return b;
  }
  int calc(int s, int t, int M) {
    int g = gcd(s, M);
    if (t % g != 0) {
      return -1;
    }
    s /= g;
    t /= g;
    M /= g;
    long long x = 0, y = 0;
    egcd(s, M, x, y);
    x = (x * t) % M;
    if (x < 0) {
      x += M;
    }
    return x;
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution sol;
  int k;
  int n, m;
  cin >> k;
  vector<int> coes(k);
  for (int i = 0; i < k; ++i) {
    cin >> coes[i];
  }
  cin >> n >> m;
  cout << sol.solve(coes, k, n, m) << "\n";
  return 0;
}