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 PYTHON):

Solution:
from __future__ import division
from sys import stdin, stdout
from math import sqrt, ceil

pow(1, 2, 3)


def write(x):
    stdout.write(str(x) + "\n")


def mmul(a, b, p):
    res = [0] * (k * k)
    for r in xrange(k):
        for i in xrange(k):
            arki = a[r * k + i]
            for c in xrange(k):
                res[r * k + c] = (res[r * k + c] + arki * b[i * k + c]) % p
    return res


def mpow(a, x, p):
    if x == 1:
        return a
    if x % 2 == 1:
        return mmul(mpow(a, x - 1, p), a, p)
    else:
        res = mpow(a, x // 2, p)
        return mmul(res, res, p)


def xgcd(b, a):
    x0, x1, y0, y1 = 1, 0, 0, 1
    while a != 0:
        q, b, a = b // a, a, b % a
        x0, x1 = x1, x0 - q * x1
        y0, y1 = y1, y0 - q * y1
    return b, x0, y0


def solve(a, c, m):
    res, x0, y0 = xgcd(a, m)
    if c % res != 0:
        return -1
    a //= res
    c //= res
    m //= res
    res, x0, y0 = xgcd(a, m)
    assert res == 1
    return (x0 * c) % m


def shanks(alpha, beta, p):
    t = int(ceil(sqrt(p)))
    small = {beta: 0}
    b = beta
    for i in range(1, t + 1):
        b = (b * alpha) % p
        small[b] = i
    for i in range(0, p, t):
        temp = pow(alpha, i, p)
        if temp in small:
            return (i - small[temp]) % (p - 1)
    assert False


k = int(stdin.readline())
b = map(int, stdin.readline().split())
assert len(b) == k
n, m = map(int, stdin.readline().split())
p = 998244353

h_n = shanks(3, m, p)

trans = b + [0] * (k * (k - 1))

for i in xrange(k - 1):
    trans[(i + 1) * k + i] = 1
trans = mpow(trans, n - k, p - 1)
c = trans[0]

h_k = solve(c, h_n, p - 1)

if h_k == -1:
    print -1
else:
    print pow(3, h_k, p)