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 p = 998244353, g = 3;
int B[105];
unordered_map<long long, long long> mp;
struct mat {
  int a[105][105];
  int n, m;
  void init0() { memset(a, 0, sizeof(a)); }
  void init1() {
    memset(a, 0, sizeof(a));
    for (int i = 1; i <= min(n, m); i++) a[i][i] = 1;
  }
};
mat operator*(mat a, mat b) {
  mat ret;
  ret.n = a.n, ret.m = b.m;
  ret.init0();
  for (int i = 1; i <= a.n; i++) {
    for (int j = 1; j <= b.m; j++) {
      for (int k = 1; k <= a.m; k++)
        ret.a[i][j] = (ret.a[i][j] + 1ll * a.a[i][k] * b.a[k][j]) % (p - 1);
    }
  }
  return ret;
}
int exgcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  int gcd = exgcd(b, a % b, x, y);
  int y2 = y, x2 = x;
  x = y2, y = x2 - a / b * y2;
  return gcd;
}
int BSGS(int b) {
  mp.clear();
  int t = 1, v = 1, x, y, m = ceil(sqrt(p));
  for (int i = 0; i < m; i++) mp[t] = i, t = 1ll * t * g % p;
  for (int i = 0; i < m; i++) {
    int gcd = exgcd(v, p, x, y);
    x = ((1ll * x * b / gcd) % p + p) % p;
    if (mp[x]) return i * m + mp[x];
    v = 1ll * v * t % p;
  }
  return -1;
}
mat kissme(mat a, int x) {
  mat ret;
  ret.n = a.n, ret.m = a.m;
  ret.init1();
  for (; x; x >>= 1) {
    if (x & 1) ret = ret * a;
    a = a * a;
  }
  return ret;
}
int kissme(int a, int x) {
  int ret = 1;
  for (; x; x >>= 1) {
    if (x & 1) ret = 1ll * ret * a % p;
    a = 1ll * a * a % p;
  }
  return ret;
}
int geta(int n, int k) {
  mat t;
  t.n = k, t.m = k;
  t.init0();
  for (int i = 1; i <= k; i++) t.a[1][i] = B[i];
  for (int i = 2; i <= k; i++) t.a[i][i - 1] = 1;
  t = kissme(t, n - k);
  return t.a[1][1];
}
int solve(int n, int b, int k) {
  int a = geta(n, k), t1 = BSGS(b), x, y;
  int gcd = exgcd(a, p - 1, x, y);
  if (t1 % gcd) return -1;
  x = ((1ll * x * t1 / gcd) % (p - 1) + (p - 1)) % (p - 1);
  return kissme(g, x);
}
int main() {
  int n, fn, k;
  scanf("%d", &k);
  for (int i = 1; i <= k; i++) scanf("%d", &B[i]);
  scanf("%d%d", &n, &fn);
  printf("%d\n", solve(n, fn, k));
  return 0;
}