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 = 700 + 10, mod = 998244353, g = 3;
inline long long rd() {
  long long x = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * w;
}
map<int, int> ha;
inline long long fpow(long long a, long long b) {
  long long an = 1;
  while (b) {
    if (b & 1) an = 1ll * an * a % mod;
    a = 1ll * a * a % mod, b >>= 1;
  }
  return an;
}
inline long long bsgs(long long a, long long b) {
  ha.clear();
  long long kk = sqrt(mod) + 1, aa = b, ka = fpow(a, kk);
  for (int i = 0; i < kk; ++i) ha[aa] = i, aa = aa * a % mod;
  aa = 1;
  for (int i = 0; i <= kk; ++i) {
    if (ha.find(aa) != ha.end() && i * kk - ha[aa] >= 0) return i * kk - ha[aa];
    aa = aa * ka % mod;
  }
  return -1;
}
int n, m, kk, b[110];
struct martix {
  int n, m;
  int a[110][110];
  martix() {}
  inline void init() {
    for (int i = 1; i <= n; i++) a[i][i] = 1;
  }
  martix(int n, int m) : n(n), m(m) {
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) a[i][j] = 0;
  }
  martix operator*(const martix &b) const {
    martix an(n, b.m);
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++)
        for (int k = 1; k <= b.m; k++)
          an.a[i][j] =
              (an.a[i][j] + 1ll * a[i][k] * b.a[k][j] % (mod - 1)) % (mod - 1);
    return an;
  }
  martix operator^(const long long &bb) const {
    martix a = *this, an(a.n, a.m);
    an.init();
    long long b = bb;
    while (b) {
      if (b & 1) an = an * a;
      a = a * a, b >>= 1;
    }
    return an;
  }
} ma, mb;
inline void exgcd(long long a, long long b, long long &d, long long &x,
                  long long &y) {
  if (!b) {
    d = a, x = 1, y = 0;
    return;
  }
  exgcd(b, a % b, d, y, x), y -= a / b * x;
}
int main() {
  n = rd();
  for (int i = 1; i <= n; ++i) b[i] = rd();
  kk = rd(), m = rd();
  ma.n = ma.m = mb.n = mb.m = n;
  ma.a[1][n] = 1;
  for (int i = 1; i < n; ++i) mb.a[i + 1][i] = 1;
  for (int i = 1; i <= n; ++i) mb.a[n - i + 1][n] = b[i] % (mod - 1);
  ma = ma * (mb ^ (kk - n));
  long long bb = bsgs(g, m), x, y, d;
  if (bb == -1) return puts("-1"), 0;
  exgcd(ma.a[1][n], mod - 1, d, x, y);
  if (bb % d) return puts("-1"), 0;
  y = (mod - 1) / d;
  x = (1ll * (x % y + y) % y * (bb / d) % y + y) % y;
  printf("%d\n", (int)fpow(g, x));
  return 0;
}