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;
struct P {
  int x, y, z;
  bool operator<(const P &a) const {
    if (x != a.x) return x > a.x;
    return y > a.y;
  }
};
vector<int> v;
int i, n, d, m, k, a, b, c;
long long o[311][111];
long long l[111][111];
int j[111111];
int e;
int dx[10] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[10] = {1, 0, -1, 0, 1, -1, 1, -1},
    dz[10] = {0, 0, 0, 0, 1, -1};
long long mod = 998244353, mod2 = 1000000009, mod3 = 2017;
long long x, z, y;
double pi = 3.14159265;
P u[1];
stack<int> s;
queue<int> q;
char r[611];
map<int, int> p;
bool as(P a, P b) { return a.x + a.y > b.x + b.y; }
long long g(long long x, long long y) {
  long long p = 1;
  for (; y; y /= 2, x *= x, x %= mod)
    if (y % 2) p *= x, p %= mod;
  return p;
}
void f(int n) {
  for (; n; n /= 2) {
    if (n % 2) {
      long long p[111][111];
      for (int t = 1; t <= a; t++)
        for (int w = 1; w <= a; w++) p[t][w] = 0;
      for (int h = 1; h <= a; h++)
        for (int t = 1; t <= a; t++)
          for (int w = 1; w <= a; w++)
            p[t][w] += o[t][h] * l[h][w], p[t][w] %= mod;
      for (int t = 1; t <= a; t++)
        for (int w = 1; w <= a; w++) l[t][w] = p[t][w];
    }
    long long p[111][111];
    for (int t = 1; t <= a; t++)
      for (int w = 1; w <= a; w++) p[t][w] = 0;
    for (int h = 1; h <= a; h++)
      for (int t = 1; t <= a; t++)
        for (int w = 1; w <= a; w++)
          p[t][w] += o[t][h] * o[h][w], p[t][w] %= mod;
    for (int t = 1; t <= a; t++)
      for (int w = 1; w <= a; w++) o[t][w] = p[t][w];
  }
}
int main() {
  scanf("%d", &a);
  for (int t = 1; t <= a; l[t][t] = 1, t++) scanf("%lld", &o[a - t + 1][a]);
  for (int t = 1; t <= a; t++) o[t][t - 1] = 1;
  scanf("%d %d", &n, &m);
  mod--;
  f(n - a);
  mod++;
  z = g(3, l[a][a]);
  k = sqrt(mod) + 1;
  x = g(z, k);
  for (long long t = 0, n = 1; t <= k; t++, n *= z, n %= mod) p[n] = t + 1;
  z = g(x, mod - 2);
  d = -1;
  for (long long t = 0, n = (long long)m % mod; t <= k; t++, n *= z, n %= mod)
    if (p[n]) {
      c = t;
      d = p[n] - 1;
      break;
    }
  if (d >= 0)
    printf("%lld", g(3, (long long)c * k + d));
  else
    puts("-1");
}