# 1.分解质因数
# 2.高精度乘法

N = 5010
primes, cnt = [0] * N, 0
st = [False] * N
sum = [0]* N

def get_primes(n):
    global cnt
    for i in range(2,n+1):
        if not st[i]:
            primes[cnt] = i
            cnt += 1
        j = 0
        while primes[j] <= n // i :
            st[primes[j]*i] = True
            if i % primes[j] == 0: break
            j +=1


#求n包含p的个数
def get(n,p):
    res = 0
    while n:
        res += n // p
        n //= p

    return res

#高精度乘法
def mul(a,b):
    c = []
    t = 0
    for i in range(len(a)):
        t += a[i] * b
        c.append(t%10)
        t //= 10

    while t:
        c.append(t%10)
        t//=10

    return c

if __name__ == '__main__':
    a,b = map(int,input().split())

    get_primes(a)

    #枚举每个质数的次数
    for i in range(cnt):
        p = primes[i]
        sum[i] = get(a,p) - get(b,p) - get(a-b,p)

    res = [1]
    for i in range(cnt):
        for j in range(sum[i]):
            res = mul(res,primes[i])

    for i in range(len(res)-1,-1,-1):
        print(res[i],end="")