CodeForces 151C Win or Freeze（簡單題）

題目鏈接

C. Win or Freeze

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer q. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.

The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.

Input

The first line contains the only integer q (1 ≤ q ≤ 10¹³).

Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.

Output

In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.

Sample test(s)

Input

6

Output

2

Input

30

Output

1
6

Input

1

Output

1
0

Note

Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory.

題意：桌子上有一個數，兩個人輪流操作。每次操作可以把這個變爲改數的因子。問先手贏還是輸，如果先手贏輸出第一手操作。

題解：如果桌子上的數的質因子個數爲2則先手輸，否則先手贏。（特判1），用sqrt（n）的方法可做。

代碼如下：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<string.h>
#include<string>
#include<stdlib.h>
typedef __int64 LL;
typedef unsigned __int64 LLU;
const int nn=11000000;
const int inf=0x3fffffff;
const LL inf64=(LL)inf*inf;
using namespace std;
LL x;
LL solve()
{
    LL i;
    LL p=x;
    int num=0;
    LL tem=1;
    for(i=2;i*i<=x;i++)
    {
        while(p%i==0)
        {
            p/=i;
            num++;
            if(num<=2)
                tem*=i;
            if(num>2)
                return tem;
        }
    }
    if(p>1)
    {
        num++;
    }
    if(num==2)
        return -1;
    return num==1?0:tem;

}
int main()
{
    while(scanf("%I64d",&x)!=EOF)
    {
        if(x==1)
        {
            puts("1");
            puts("0");
            continue;
        }
        LL ix=solve();
        if(ix==-1)
        {
            puts("2");
        }
        else
        {
            puts("1");
            printf("%I64d\n",ix);
        }
    }
    return 0;
}