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;
inline int read() {
  int p = 0, w = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') p = p * 10 + c - '0', c = getchar();
  return p * w;
}
const int sid = 105;
const int mod = 998244353;
const int g = 3;
inline int mul(int a, int b) { return 1ll * a * b % mod; }
inline int fp(int a, int k) {
  int ret = 1;
  for (; k; k >>= 1, a = mul(a, a))
    if (k & 1) ret = mul(ret, a);
  return ret;
}
int n, m, k, b[sid];
struct mtx {
  int f[sid][sid];
  mtx() {}
  mtx(int flag) {
    if (flag == 0) {
      for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++) f[i][j] = 0;
    }
    if (flag == 1) {
      for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++) f[i][j] = (i == j) ? 1 : 0;
    }
  }
  int *operator[](const int x) { return f[x]; }
  friend mtx operator*(mtx a, mtx b) {
    mtx c(0);
    for (int i = 0; i < k; i++)
      for (int j = 0; j < k; j++)
        for (int p = 0; p < k; p++)
          c[i][j] = (c[i][j] + 1ll * a[i][p] * b[p][j] % (mod - 1)) % (mod - 1);
    return c;
  }
} A, B;
inline mtx fp(mtx a, int k) {
  mtx ret(1);
  for (; k; k >>= 1, a = a * a)
    if (k & 1) ret = ret * a;
  return ret;
}
map<int, int> H;
inline int BSGS(int A, int B) {
  H.clear();
  H[1] = 0;
  int Ai = 1, Aj = 1, m = ceil(sqrt(mod));
  for (register int i = 1; i < m; i++)
    Ai = 1ll * Ai * A % mod, H[1ll * Ai * B % mod] = i;
  Ai = 1ll * Ai * A % mod;
  for (register int i = 1; i <= m; i++) {
    Aj = 1ll * Aj * Ai % mod;
    if (H[Aj]) return 1ll * i * m - H[Aj];
  }
}
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
inline void exgcd(int &x, int &y, int a, int b) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(y, x, b, a % b);
  y -= a / b * x;
}
inline void Solve() {
  A[0][k - 1] = 1;
  for (register int i = 0; i <= k - 1; i++) B[i][k - 1] = b[k - i];
  for (register int i = 0; i <= k - 2; i++) B[i + 1][i] = 1;
  A = A * fp(B, n - k);
  int p = A[0][k - 1], t = BSGS(g, m);
  if (t % gcd(p, mod - 1))
    printf("-1\n");
  else {
    int gd = gcd(p, mod - 1);
    int x, y, a = p, b = mod - 1;
    t /= gd;
    a /= gd;
    b /= gd;
    exgcd(x, y, a, b);
    x = (x + mod - 1) % (mod - 1);
    x = 1ll * x * t % (mod - 1);
    printf("%d\n", fp(g, x));
  }
}
int main() {
  k = read();
  for (register int i = 1; i <= k; i++) b[i] = read();
  n = read();
  m = read();
  Solve();
  return 0;
}