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;
template <typename T>
inline T read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x = f ? -x : x;
}
const int N = 100 + 10;
const int P = 998244353;
int K, n, m, i, j, x, y, yy, t, b[N];
struct matrix {
  int v[N][N];
  void O() { memset(v, 0, sizeof(v)); }
  void E() {
    memset(v, 0, sizeof(v));
    for (int i = 0; i < K; ++i) v[i][i] = 1;
  }
  matrix operator*(const matrix &b) const {
    matrix c;
    c.O();
    for (int k = 0; k < K; k++) {
      for (int i = 0; i < K; i++)
        if (v[i][k]) {
          for (int j = 0; j < K; j++) {
            c.v[i][j] = (c.v[i][j] + (long long)v[i][k] * b.v[k][j]) % (P - 1);
          }
        }
    }
    return c;
  }
  matrix operator^(int p) const {
    matrix y;
    y.E();
    matrix x;
    memcpy(x.v, v, sizeof(v));
    while (p) {
      if (p & 1) y = y * x;
      x = x * x;
      p >>= 1;
    }
    return y;
  }
} mat;
int fexp(int a, int n, int mod) {
  int res = 1;
  while (n) {
    if (n & 1) res = (long long)res * a % mod;
    a = (long long)a * a % mod;
    n >>= 1;
  }
  return res;
}
int baby_step_gaint_step(int a, int b, int p) {
  map<int, int> hash;
  hash.clear();
  b %= p;
  int t = (int)sqrt(p) + 1;
  for (int j = 0; j < t; ++j) {
    int val = (long long)b * fexp(a, j, p) % p;
    hash[val] = j;
  }
  a = fexp(a, t, p);
  if (a == 0) return b == 0 ? 1 : -1;
  for (int i = 0; i <= t; ++i) {
    int val = fexp(a, i, p);
    int j = hash.find(val) == hash.end() ? -1 : hash[val];
    if (j >= 0 && i * t - j >= 0) return i * t - j;
  }
  return -1;
}
int exgcd(int a, int b, int &x, int &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  int d = exgcd(b, a % b, x, y);
  int z = x;
  x = y;
  y = z - y * (a / b);
  return d;
}
int main() {
  read(K);
  for (i = 1; i <= K; ++i) read(b[i]);
  read(n), read(m);
  for (i = 0; i < K; ++i) mat.v[0][i] = b[i + 1];
  for (i = 1; i < K; ++i) mat.v[i][i - 1] = 1;
  mat = mat ^ (n - K);
  y = mat.v[0][0];
  t = baby_step_gaint_step(3, m, P);
  if (t == -1) return puts("-1"), 0;
  int d = exgcd(y, P - 1, x, yy);
  if (t % d != 0)
    puts("-1");
  else {
    x = 1LL * x * t / d % (P - 1);
    x = (x % (P - 1) + (P - 1)) % (P - 1);
    printf("%d\n", fexp(3, x, P));
  }
  return 0;
}