# 変形してQ*(M+N) = N*M*P
# gcd(P, Q)で両辺を割る
# もしP=Q=1の場合、M=N=2
# それ以外の場合、N,MはQの素因数を持つ、N+MはPの倍数
# Qの約数でNを全探索すればいい

def divisors(n):
    lower_divisors , upper_divisors = [], []
    i = 1
    while i*i <= n:
        if n % i == 0:
            lower_divisors.append(i)
            if i != n // i:
                upper_divisors.append(n//i)
        i += 1
    return lower_divisors + upper_divisors[::-1]

from math import gcd
P, Q = map(int, input().split())
g = gcd(P, Q)
P = P//g
Q = Q//g
if P == 1:
    print(1)
    print(Q*2, Q*2)
else:
    ans_list = []
    divs = divisors(Q)
    for n in divs:
        if (P*n-Q) != 0 and (n*Q)%(P*n-Q) == 0 and (n*Q)//(P*n-Q) > 0:
            if Q*(((n*Q)//(P*n-Q))+n) == P*((n*Q)//(P*n-Q))*n:
                ans_list.append((n, (n*Q)//(P*n-Q)))
    for k in range(2, 10**7):
        n_ = n*k
        if (P*n_-Q) != 0 and (n_*Q)%(P*n_-Q) == 0 and (n_*Q)//(P*n_-Q) > 0:
            if Q*(((n_*Q)//(P*n_-Q))+n_) == P*((n_*Q)//(P*n_-Q))*n_:
                ans_list.append((n_, (n_*Q)//(P*n_-Q)))        
    print(len(ans_list))
    for n, m in ans_list:
        print(n, m)