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>
void mult(int aa[][100], int bb[][100], int cc[][100], int k) {
  int h, i, j;
  for (i = 0; i < k; i++) {
    int *ci = cc[i];
    for (j = 0; j < k; j++) ci[j] = 0;
    for (h = 0; h < k; h++) {
      int *bh = bb[h];
      long long aih = aa[i][h];
      for (j = 0; j < k; j++) ci[j] = (ci[j] + aih * bh[j]) % (998244353 - 1);
    }
  }
}
void power_(int aa[][100], int pp[][100], int tt[][100], int k, int x) {
  if (x == 0) {
    int i, j;
    for (i = 0; i < k; i++)
      for (j = 0; j < k; j++) pp[i][j] = i == j;
    return;
  }
  if (x % 2 == 0) {
    power_(aa, tt, pp, k, x / 2);
    mult(tt, tt, pp, k);
  } else {
    power_(aa, pp, tt, k, x / 2);
    mult(pp, pp, tt, k);
    mult(tt, aa, pp, k);
  }
}
int power(int a, int x) {
  int p;
  if (x == 0) return 1;
  p = power(a, x / 2);
  p = (long long)p * p % 998244353;
  if (x % 2 == 1) p = (long long)p * a % 998244353;
  return p;
}
int totient(int n) {
  int a, t = n;
  for (a = 2; a * a <= n; a++)
    if (n % a == 0) {
      t = t / a * (a - 1);
      do n /= a;
      while (n % a == 0);
    }
  if (n > 1) {
    a = n;
    t = t / a * (a - 1);
  }
  return t;
}
int test(int r, int t) {
  int a, t_ = t;
  for (a = 2; a * a <= t; a++)
    if (t % a == 0) {
      if (power(r, t_ / a) == 1) return 0;
      do t /= a;
      while (t % a == 0);
    }
  if (t > 1) {
    a = t;
    if (power(r, t_ / a) == 1) return 0;
  }
  return 1;
}
int primroot(int t) {
  while (1) {
    int r = rand() % (998244353 - 2) + 2;
    if (test(r, t)) return r;
  }
}
struct S {
  int i, x;
} pp[31596], qq[31596];
int compare(const void *a, const void *b) {
  struct S *p = (struct S *)a;
  struct S *q = (struct S *)b;
  return p->x - q->x;
}
int dislog(int a, int b, int t) {
  int m = sqrt(998244353) + 1, i, j, x, am;
  x = 1;
  for (j = 0; j < m; j++) {
    struct S *s = &qq[j];
    s->i = j;
    s->x = x;
    x = (long long)x * a % 998244353;
  }
  am = power(a, t - m % t);
  x = b;
  for (i = 0; i < m; i++) {
    struct S *s = &pp[i];
    s->i = i;
    s->x = x;
    x = (long long)x * am % 998244353;
  }
  qsort(pp, m, sizeof *pp, compare);
  qsort(qq, m, sizeof *qq, compare);
  for (i = 0, j = 0; i < m && j < m;) {
    if (pp[i].x == qq[j].x) return (pp[i].i * m + qq[j].i) % t;
    if (pp[i].x < qq[j].x)
      i++;
    else
      j++;
  }
  return -1;
}
int x_, y_;
int gcd_(int a, int b) {
  if (b == 0) {
    x_ = 1;
    y_ = 0;
    return a;
  } else {
    int d = gcd_(b, a % b);
    int t = x_ - a / b * y_;
    x_ = y_;
    y_ = t;
    return d;
  }
}
int main() {
  static int bb[100], aa[100][100], pp[100][100], tt[100][100];
  int k, i, j, n, m, d, t, r, l, ans;
  srand(time(NULL));
  scanf("%d", &k);
  for (i = 0; i < k; i++) scanf("%d", &bb[i]);
  scanf("%d%d", &n, &m);
  for (i = 0; i < k - 1; i++) aa[i][i + 1] = 1;
  for (j = 0; j < k; j++) aa[k - 1][j] = bb[k - 1 - j];
  power_(aa, pp, tt, k, n - k);
  d = gcd_(pp[k - 1][k - 1], 998244353 - 1);
  t = totient(998244353);
  r = primroot(t);
  l = dislog(r, m, t);
  ans = -1;
  if (l % d == 0) {
    if (x_ < 0) x_ += 998244353 - 1;
    ans = power(r, (long long)x_ * (l / d) % t);
  }
  printf("%d\n", ans);
  return 0;
}