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 modl = 998244353;
int n, m, kk;
int coef[100][100], ini_v[100];
const int maxl_step = (int)1e5;
map<int, int> baby_pool;
int prm_r = 3;
int mpow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = ((long long)res) * ((long long)x) % modl;
    x = ((long long)x) * ((long long)x) % modl;
    y >>= 1;
  }
  return res;
}
int dis_log(int bse, int mm) {
  int i, j, k;
  k = 1;
  for (i = 0; i <= maxl_step; i++) {
    baby_pool[k] = i;
    k = ((long long)k) * ((long long)bse) % modl;
  }
  k = mpow(bse, modl - 2);
  k = mpow(k, maxl_step);
  map<int, int>::iterator it;
  for (i = 0, j = mm; i <= maxl_step; i++) {
    if ((it = baby_pool.find(j)) != baby_pool.end()) break;
    j = ((long long)j) * ((long long)k) % modl;
  }
  if (i > maxl_step)
    return -1;
  else
    return baby_pool[j] + maxl_step * i;
}
int tmp_v[100], tmp_coef[100][100];
void mul(int nn) {
  int i, j, k, a, b;
  while (nn) {
    if (nn & 1) {
      for (i = 0; i < kk; i++) {
        tmp_v[i] = 0;
        for (j = 0; j < kk; j++) {
          k = ((long long)coef[i][j]) * ((long long)ini_v[j]) % (modl - 1);
          tmp_v[i] = (((long long)tmp_v[i]) + ((long long)k)) % (modl - 1);
        }
      }
      for (i = 0; i < kk; i++) ini_v[i] = tmp_v[i];
    }
    for (i = 0; i < kk; i++) {
      for (j = 0; j < kk; j++) {
        tmp_coef[i][j] = 0;
        for (k = 0; k < kk; k++) {
          a = ((long long)coef[i][k]) * ((long long)coef[k][j]) % (modl - 1);
          tmp_coef[i][j] =
              (((long long)tmp_coef[i][j]) + ((long long)a)) % (modl - 1);
        }
      }
    }
    for (i = 0; i < kk; i++)
      for (j = 0; j < kk; j++) coef[i][j] = tmp_coef[i][j];
    nn >>= 1;
  }
}
int main() {
  int i, j, k;
  cin >> kk;
  for (i = 1; i < kk; i++) coef[i - 1][i] = 1;
  for (i = 0; i < kk; i++) cin >> coef[kk - 1][kk - i - 1];
  ini_v[kk - 1] = 1;
  cin >> n >> m;
  mul(n - 1);
  k = dis_log(mpow(prm_r, ini_v[0]), m);
  if (k == -1)
    printf("-1\n");
  else
    printf("%d\n", mpow(prm_r, k));
  return 0;
}