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 N = 100;
const int G = 3;
const int P = 998244353;
long long quickpow(long long x, long long y) {
  long long cur = x, ret = 1ll;
  for (int i = 0; y; i++) {
    if (y & (1ll << i)) {
      y -= (1ll << i);
      ret = ret * cur % P;
    }
    cur = cur * cur % P;
  }
  return ret;
}
struct Matrix {
  long long a[N + 3][N + 3];
  int sz1, sz2, sz;
  void init() {
    for (int i = 1; i <= sz1; i++)
      for (int j = 1; j <= sz2; j++) a[i][j] = 0ll;
  }
  Matrix() {}
  Matrix(int _sz) {
    sz = sz1 = sz2 = _sz;
    init();
  }
  Matrix(int _sz1, int _sz2) {
    sz1 = _sz1, sz2 = _sz2;
    init();
  }
  void uinit(int _sz) {
    sz = sz1 = sz2 = _sz;
    for (int i = 1; i <= sz; i++)
      for (int j = 1; j <= sz; j++) a[i][j] = (i == j) ? 1 : 0;
  }
  void output() {
    for (int i = 1; i <= sz1; i++) {
      for (int j = 1; j <= sz2; j++) printf("%lld ", a[i][j]);
      puts("");
    }
  }
};
Matrix operator*(Matrix x, Matrix y) {
  Matrix ret = Matrix(x.sz1, y.sz2);
  for (int i = 1; i <= x.sz1; i++) {
    for (int j = 1; j <= x.sz2; j++) {
      for (int k = 1; k <= y.sz2; k++) {
        ret.a[i][j] = (ret.a[i][j] + x.a[i][k] * y.a[k][j]) % (P - 1ll);
      }
    }
  }
  return ret;
}
Matrix mquickpow(Matrix x, long long y) {
  Matrix cur = x, ret;
  ret.uinit(x.sz);
  for (int i = 0; y; i++) {
    if (y & (1ll << i)) {
      y -= (1ll << i);
      ret = ret * cur;
    }
    cur = cur * cur;
  }
  return ret;
}
namespace BSGS {
const int B = 31595;
map<long long, int> mp;
void init() {
  long long bs = quickpow(G, B);
  long long j = 1ll;
  for (int i = 0; i <= P; i += B, j = (j * bs) % P) {
    mp[j] = i;
  }
}
long long Logarithm(long long x) {
  long long j = 1ll;
  for (int i = 0; i <= B; i++, j = (j * G) % P) {
    long long tmp = x * j % P;
    if (mp.count(tmp)) {
      long long ret = (mp[tmp] - i + (P - 1)) % (P - 1);
      return ret;
    }
  }
  return P - 1;
}
}  // namespace BSGS
Matrix mA, mB, mC;
long long a[N + 3], b[N + 3];
int n;
long long m, p, q, lq, lx;
long long gcd(long long x, long long y) { return y == 0 ? x : gcd(y, x % y); }
void exgcd(long long _a, long long _b, long long &_x, long long &_y) {
  if (_b == 0ll) {
    _x = 1ll, _y = 0ll;
    return;
  }
  exgcd(_b, _a % _b, _x, _y);
  long long tmp = _x;
  _x = _y;
  _y = tmp - (_a / _b) * _y;
}
long long CongruenceEquation(long long _a, long long _b, long long mod) {
  long long g = gcd(_a, mod), x, y;
  exgcd(_a / g, mod / g, x, y);
  return (x * (_b / g) % mod + mod) % mod;
}
int main() {
  BSGS::init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &b[i]);
  scanf("%I64d", &m);
  mA = Matrix(1, n);
  mA.a[1][1] = 1ll;
  mB = Matrix(n);
  for (int i = 1; i < n; i++) mB.a[i][i + 1] = 1ll;
  for (int i = 1; i <= n; i++) mB.a[i][1] = b[i];
  mC = mA * mquickpow(mB, m - n);
  p = mC.a[1][1];
  scanf("%I64d", &q);
  lq = BSGS::Logarithm(q);
  if (lq % gcd(P - 1, p) != 0) {
    printf("-1\n");
    return 0;
  }
  lx = CongruenceEquation(p, lq, P - 1);
  long long ans = quickpow(G, lx);
  printf("%I64d\n", ans);
  return 0;
}