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 = 209;
const int mod = 998244353;
const int pmod = mod - 1;
int qpow(int x, int y) {
  int res = 1;
  for (; y; x = 1ll * x * x % mod, y >>= 1)
    if (y & 1) res = 1ll * res * x % mod;
  return res;
}
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
int n, m, k;
struct Martix {
  int G[N][N];
  void clear() {
    for (int i = 1, iE = m; i <= iE; i++)
      for (int j = 1, jE = m; j <= jE; j++) G[i][j] = 0;
  }
  void start() {
    clear();
    for (int i = 1, iE = m; i <= iE; i++) G[i][i] = 1;
  }
} P;
Martix operator*(Martix aa, Martix bb) {
  Martix res;
  res.clear();
  for (int i = 1, iE = m; i <= iE; i++)
    for (int j = 1, jE = m; j <= jE; j++)
      for (int k = 1, kE = m; k <= kE; k++)
        (res.G[i][j] += 1ll * aa.G[i][k] * bb.G[k][j] % pmod) %= pmod;
  return res;
}
Martix operator^(Martix aa, int bb) {
  Martix res;
  res.start();
  for (; bb; aa = aa * aa, bb >>= 1)
    if (bb & 1) res = res * aa;
  return res;
}
map<int, int> mp;
int BSGS(int x, int y) {
  int B = sqrt(mod) + 1, now = y, res = 1, Pow = qpow(x, B);
  mp[now] = 0;
  for (int i = 1, iE = B - 1; i <= iE; i++)
    now = 1ll * now * x % mod, mp[now] = i;
  for (int i = 1, iE = B; i <= iE; i++) {
    res = 1ll * res * Pow % mod;
    if (mp[res]) return i * B - mp[res];
  }
  return -1;
}
void exgcd(int a, int b, long long &x, long long &y) {
  if (b == 0) return x = 1, void();
  exgcd(b, a % b, y, x);
  y -= (a / b) * x;
  long long tt = x / b;
  x -= tt * b, y += tt * a;
}
int NY(int a, int Mod) {
  long long x, y;
  exgcd(a, Mod, x, y);
  return (x % Mod + Mod) % Mod;
}
int main() {
  scanf("%d", &m);
  for (int i = 1, iE = m; i <= iE; i++) scanf("%d", &P.G[m - i + 1][m]);
  for (int i = 2, iE = m; i <= iE; i++) P.G[i][i - 1] = 1;
  scanf("%d%d", &n, &k);
  if (k == 1) return puts("1"), 0;
  int cs = (P ^ (n - m)).G[m][m];
  int YG = BSGS(3, k);
  if (YG == -1) return puts("-1"), 0;
  int mm = gcd(YG, cs);
  YG /= mm, cs /= mm;
  if (gcd(mod - 1, cs) != 1) return puts("-1"), 0;
  printf("%d\n", qpow(3, 1ll * YG * NY(cs, mod - 1) % (mod - 1)));
  return 0;
}