1106_F. Lunar New Year and a Recursive Sequence
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.
```python
from math import ceil, sqrt

p = 998244353

def bsgs(g, h):
    '''
    Solve for x in h = g^x mod p given a prime p.
    '''
    N = ceil(sqrt(p - 1))  # phi(p) is p-1 if p is prime

    # Store hashmap of g^{1...m} (mod p). Baby step.
    tbl = {pow(g, i, p): i for i in range(N)}

    # Precompute via Fermat's Little Theorem
    c = pow(g, N * (p - 2), p)

    # Search for an equivalence in the table. Giant step.
    for j in range(N):
        y = (h * pow(c, j, p)) % p
        if y in tbl:
            return j * N + tbl[y]

def gcd(a, b):
    return b if a % b == 0 else gcd(b, a % b)

def xgcd(a, b):
    """return (g, x, y) such that a*x + b*y = g = gcd(x, y)"""
    x0, x1, y0, y1 = 0, 1, 1, 0
    while a != 0:
        q, b, a = b // a, a, b % a
        y0, y1 = y1, y0 - q * y1
        x0, x1 = x1, x0 - q * x1
    return b, x0, y0

def inv(a, m):
    g, x, y = xgcd(a, m)
    if g != 1:
        return None
    return (x + m) % m

# solve a = bx (mod m)
def div(a, b, m):
    k = gcd(b, m)
    if a % k != 0:
        return None

    ak = a // k
    bk = b // k
    mk = m // k
    inv_bk = inv(bk, mk)
    return (ak * inv_bk) % mk

def matmul(A, B):
    m = len(A)
    C = [[0 for _ in range(m)] for _ in range(m)]
    for i in range(m):
        for k in range(m):
            for j in range(m):
                C[i][j] += A[i][k]*B[k][j] % (p-1) 
    for i in range(m):
        for j in range(m):
            C[i][j] %= (p-1) 
    return C

def Id(n):
    return [[int(i == j) for i in range(n)] for j in range(n)]

def matexp(A, n):
    if n == 0:
        return Id(len(A))
    h = matexp(A, n//2)

    R = matmul(h, h)
    if n % 2 == 1:
        R = matmul(A, R)
    return R

def solve():
    k = int(input())

    A = [[0 for _ in range(k)] for _ in range(k)]
    A[0] = [int(x) for x in input().split()]
    for i in range(k-1):
        A[i+1][i] = 1

    n, v = [int(x) for x in input().split()]

    e = matexp(A, n-k)[0][0]
    g = 3
    u = bsgs(g, v) 

    x = div(u, e, p-1)
    if x is not None:
        print(pow(g, x, p))
    else:
        print(-1)

if __name__ == '__main__':
    solve()

```