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 long long maxn = 1e5 + 10;
const int mod = 998244353;
struct M {
  long long d[105][105];
  int size;
  M operator*(M t) {
    M tt;
    tt.size = t.size;
    memset(tt.d, 0, sizeof d);
    for (int i = 0; i < size; i++)
      for (int j = 0; j < size; j++) {
        for (int k = 0; k < size; k++) {
          tt.d[i][j] =
              (tt.d[i][j] + (d[i][k] * t.d[k][j]) % (mod - 1)) % (mod - 1);
        }
      }
    return tt;
  }
};
M quick(M t, long long n) {
  M tt;
  tt.size = t.size;
  for (int i = 0; i < tt.size; i++)
    for (int j = 0; j < tt.size; j++) tt.d[i][j] = 0;
  for (int i = 0; i < 105; i++) tt.d[i][i] = 1;
  while (n) {
    if (n & 1) tt = tt * t;
    t = t * t;
    n >>= 1;
  }
  return tt;
}
long long quick(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
bool notpri[maxn];
int pri[maxn], zyz[maxn];
void shai(int N) {
  notpri[1] = 1;
  for (int i = 2; i <= N; ++i) {
    if (!notpri[i]) pri[++pri[0]] = i;
    for (int j = 1; j <= pri[0] && (long long)i * pri[j] <= N; j++) {
      notpri[i * pri[j]] = 1;
      if (i % pri[j] == 0) break;
    }
  }
}
int find_root(int x) {
  int tmp = x - 1;
  for (int i = 1; tmp && i <= pri[0]; i++) {
    if (tmp % pri[i] == 0) {
      zyz[++zyz[0]] = pri[i];
      while (tmp % pri[i] == 0) tmp /= pri[i];
    }
  }
  for (int g = 2; g <= x - 1; g++) {
    bool flag = 1;
    for (int i = 1; i <= zyz[0]; i++) {
      if (quick(g, (x - 1) / zyz[i], x) == 1) {
        flag = 0;
        break;
      }
    }
    if (flag) return g;
  }
  return 1;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    long long x1, y1;
    long long d = exgcd(b, a % b, x1, y1);
    x = y1;
    y = x1 - a / b * y1;
    return d;
  }
}
int main() {
  shai(maxn);
  M t;
  int k;
  cin >> k;
  t.size = k;
  for (int i = 0; i < k; i++)
    for (int j = 0; j < k; j++) t.d[i][j] = 0;
  for (int i = 0; i < k; i++) {
    cin >> t.d[k - 1 - i][k - 1];
  }
  for (int i = 0; i < k - 1; i++) {
    t.d[i + 1][i] = 1;
  }
  int n, m;
  cin >> n >> m;
  t = quick(t, n - k);
  long long x = t.d[k - 1][k - 1];
  long long g = 3;
  long long M = sqrt((double)mod), gg = 1, mm = m, inv;
  map<long long, long long> num;
  map<long long, bool> app;
  app[1] = 1;
  num[1] = 0;
  for (int i = 1; i <= M - 1; i++) {
    gg = gg * g % mod;
    if (app[gg] == 0) {
      app[gg] = 1;
      num[gg] = i;
    }
  }
  gg = gg * g % mod;
  inv = quick(gg, mod - 2, mod);
  long long ans;
  for (int i = 0; i <= M; i++) {
    if (app[mm]) {
      ans = i * M + num[mm];
      break;
    }
    mm = mm * inv % mod;
  }
  long long t1, t2;
  long long gcd = exgcd(x, mod - 1, t1, t2);
  if (ans % gcd != 0) {
    cout << -1;
    return 0;
  }
  inv = ((ans / gcd * t1) % (mod - 1) + mod - 1) % (mod - 1);
  cout << quick(g, inv, mod);
}