# 素因数分解　'26/02/14改訂
# def PF(n)内のxを素数に限定しないと大量処理では時間がかかる

def eratos(n):
    primes = [True] * (n+1)
    primes[0], primes[1] = False, False
    for i in range(2, int(n**0.5)+1):
        if primes[i]:
            for j in range(i**2, n+1, i):
                primes[j] = False
    ret = [num for num, is_prime in enumerate(primes) if is_prime]
    return ret

prime = eratos(10 ** 4 +1)

def PF(n):
    y = n
    p = []
    for x in prime:
        if x * x > n:
            break
        cnt = 0
        while y % x == 0:
            y //= x
            cnt += 1
        if cnt != 0:
            p.append((x,cnt))
    if y != 1:
        p.append((y,1))
    return p