'''
利用快速幂求x^x mod 1000 的数值n
利用隔板法，最后答案就是组合数C(n-1, k-1)
(n-1个空隙中选出位置不同的k-1个)
'''
def pow_mod(a, k, p):
    t = []
    pow_val = 1             # 2的次幂数, 初始是2^0 = 1
    a_pow = a % p           # a^(2 ^ i)的数值, 初始是a^(2^0) = a
    while pow_val <= k:
        t.append(a_pow)
        a_pow = (a_pow*a_pow) % p
        pow_val <<= 1

    ans = 1
    for i in range(len(t)):
        if k & 1:
            ans = (ans * t[i]) % p
        k >>= 1
    return ans


import sys
sys.setrecursionlimit(999999)


from functools import lru_cache
@lru_cache(typed=False, maxsize=128000000)
def conbination_num(n, m):
    if m == n:
        return 1

    if m == 0:
        return 1

    if m == 1:
        return n

    return conbination_num(n-1, m-1) + conbination_num(n-1, m)


k, x = map(int, input().split())
n = pow_mod(x, x, 1000)

print(conbination_num(n-1, k-1))