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>
const int N = 105;
const int MOD = 998244353;
int k, n, m;
std::map<int, int> w;
struct Matrix {
  int a[N][N];
} a, b, c;
int ksm(int x, int y, int p) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = (long long)ans * x % p;
    x = (long long)x * x % p;
    y >>= 1;
  }
  return ans;
}
int solve(int p, int b, int n) {
  int ny = ksm(b, p - 2, p), m = sqrt(p);
  w.clear();
  int v = n;
  w[n] = -1;
  for (int i = 1; i < m; i++) {
    v = (long long)v * ny % p;
    if (!w[v]) w[v] = i;
  }
  b = ksm(b, m, p);
  v = 1;
  for (int i = 0; i <= m; i++) {
    int x = w[v];
    if (x) {
      if (x == -1)
        return i * m;
      else
        return i * m + x;
    }
    v = (long long)v * b % p;
  }
  return -1;
}
int BSGS(int p, int b, int n) { return solve(p, b, n); }
void mul1() {
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= k; j++) c.a[i][j] = 0;
  for (int l = 1; l <= k; l++)
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= k; j++)
        (c.a[i][j] += (long long)a.a[i][l] * b.a[l][j] % (MOD - 1)) %=
            (MOD - 1);
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= k; j++) b.a[i][j] = c.a[i][j];
}
void mul2() {
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= k; j++) c.a[i][j] = 0;
  for (int l = 1; l <= k; l++)
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= k; j++)
        (c.a[i][j] += (long long)a.a[i][l] * a.a[l][j] % (MOD - 1)) %=
            (MOD - 1);
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= k; j++) a.a[i][j] = c.a[i][j];
}
void ksm(int y) {
  for (int i = 1; i <= k; i++) b.a[i][i] = 1;
  while (y) {
    if (y & 1) mul1();
    y >>= 1;
    mul2();
  }
}
int gcd(int x, int y) {
  if (!y)
    return x;
  else
    return gcd(y, x % y);
}
void exgcd(int x, int y, int &a, int &b) {
  if (!y) {
    a = 1;
    b = 0;
    return;
  }
  exgcd(y, x % y, a, b);
  int tmp = a;
  a = b;
  b = tmp - (x / y) * b;
}
int main() {
  scanf("%d", &k);
  for (int i = 1; i <= k; i++) scanf("%d", &a.a[k - i + 1][k]);
  scanf("%d%d", &n, &m);
  if (m == 1) {
    puts("1");
    return 0;
  }
  for (int i = 1; i < k; i++) a.a[i + 1][i] = 1;
  ksm(n - k);
  int u = b.a[k][k], y = BSGS(MOD, 3, m), p = MOD - 1;
  int d = gcd(u, p);
  if (y % d > 0) {
    puts("-1");
    return 0;
  }
  int t1, t2;
  exgcd(u, p, t1, t2);
  t1 = ((long long)t1 * (y / d) % p + p) % p;
  printf("%d\n", ksm(3, t1, MOD));
  return 0;
}