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 mod = 998244353;
const long long P = 998244352;
const int N = 100 + 5;
long long k, n, m;
long long b[N];
struct N_time_remaining {
  static const int N = 100100;
  long long fpow(long long a, long long b, long long p) {
    long long res = 1;
    a %= p;
    for (; b; b >>= 1, a = a * a % p)
      if (b & 1) res = res * a % p;
    return res;
  }
  long long ex_gcd(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
      x = 1, y = 0;
      return a;
    } else {
      long long r = ex_gcd(b, a % b, y, x);
      y -= x * (a / b);
      return r;
    }
  }
  vector<long long> a;
  bool check(long long g, long long p) {
    int sz = a.size();
    for (int i = 0; i < sz; i++)
      if (fpow(g, (p - 1) / a[i], p) == 1) return 0;
    return 1;
  }
  long long primitive_root(long long p) {
    long long tmp = p - 1;
    for (int i = 2; i <= tmp / i; i++) {
      if (tmp % i == 0) {
        a.push_back(i);
        while (tmp % i == 0) tmp /= i;
      }
    }
    if (tmp != 1) a.push_back(tmp);
    long long g = 1;
    while (true) {
      if (check(g, p)) return g;
      ++g;
    }
  }
  struct sa {
    long long x;
    int id;
    bool operator<(const sa &b) const {
      if (x == b.x) return id < b.id;
      return x < b.x;
    }
  } rec[N];
  long long discerte_log(long long x, long long n, long long m) {
    int s = (int)(sqrt((double)m + 0.5));
    while ((long long)s * s <= m) s++;
    long long cur = 1;
    sa tmp;
    for (int i = 0; i < s; i++) {
      tmp.x = cur;
      tmp.id = i;
      rec[i] = tmp;
      cur = cur * x % m;
    }
    sort(rec, rec + s);
    long long mul = fpow(cur, m - 2, m) % m;
    cur = 1;
    for (int i = 0; i < s; i++) {
      long long more = n * cur % m;
      tmp.x = more;
      tmp.id = -1;
      int j = (int)(lower_bound(rec, rec + s, tmp) - rec);
      if (rec[j].x == more) return i * s + rec[j].id;
      cur = cur * mul % m;
    }
    return -1;
  }
  vector<long long> residue(long long p, long long n, long long a) {
    vector<long long> ret;
    if (a == 0) {
      ret.push_back(0);
      return ret;
    }
    if (n == 0) {
      if (a == 1) ret.push_back(1);
      return ret;
    }
    long long g = primitive_root(p);
    long long m = discerte_log(g, a, p);
    if (m == -1) return ret;
    long long A = n, B = p - 1, C = m, x, y;
    long long G = ex_gcd(A, B, x, y);
    if (C % G != 0) return ret;
    x = x * (C / G) % B;
    long long delta = B / G;
    for (int i = 0; i < G; i++) {
      x = ((x + delta) % B + B) % B;
      ret.push_back(fpow(g, x, p));
    }
    sort(ret.begin(), ret.end());
    ret.erase(unique(ret.begin(), ret.end()), ret.end());
    return ret;
  }
} rt;
struct Matrix {
  static const int maxn = 105;
  long long mp[maxn][maxn];
  void getUnit(long long sz) {
    for (int i = 1; i <= sz; i++)
      for (int j = 1; j <= sz; j++) mp[i][j] = 0;
    for (int i = 1; i <= sz; i++) mp[i][i] = 1LL;
  }
};
Matrix mul(Matrix a, Matrix b, long long sz) {
  Matrix res;
  for (int i = 1; i <= sz; ++i) {
    for (int j = 1; j <= sz; ++j) {
      res.mp[i][j] = 0LL;
      for (int k = 1; k <= sz; ++k) {
        res.mp[i][j] += a.mp[i][k] * b.mp[k][j] % P;
        res.mp[i][j] %= P;
      }
    }
  }
  return res;
}
Matrix pw(Matrix a, long long b, long long sz) {
  Matrix res;
  res.getUnit(sz);
  while (b) {
    if (b & 1) res = mul(res, a, sz);
    a = mul(a, a, sz);
    b >>= 1;
  }
  return res;
}
long long calc() {
  Matrix bs, p;
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= k; j++) bs.mp[i][j] = p.mp[i][j] = 0;
  for (int i = 1; i <= k; i++) bs.mp[1][i] = b[i];
  for (int i = 2; i <= k; i++) bs.mp[i][i - 1] = 1;
  p.mp[1][1] = 1;
  bs = pw(bs, n - k, k);
  bs = mul(bs, p, k);
  return bs.mp[1][1] % P;
}
int main() {
  scanf("%lld", &k);
  for (int i = 1; i <= k; i++) scanf("%lld", &b[i]);
  scanf("%lld%lld", &n, &m);
  long long Pn = calc();
  vector<long long> v = rt.residue(mod, Pn, m);
  int sz = v.size();
  printf("%lld\n", sz > 0 ? v[0] : -1LL);
  return 0;
}