Description

Given a positive integer X, an X-factor chain of length m is a sequence of integers,

1 = X₀, X₁, X₂, …,X_m= X

satisfying

X_i < X_i+1 and X_i |X_i+1 where a | b means a perfectly divides intob.

Now we are interested in the maximum length of X-factor chains and the number of chains of such length.

Input

The input consists of several test cases. Each contains a positive integer X (X ≤ 2²⁰).

Output

For each test case, output the maximum length and the number of such X-factors chains.

Sample Input

2
3
4
10
100

Sample Output

1 1
1 1
2 1
2 2
4 6

題意：給你一個數X，將X分解成1~X的因子數列，前一個數可以整數後一個數，求滿足條件的最大鏈長以及有多少條這樣長的鏈。

分析：很容易想到了素因數分解。不難推出，我們要求的最大鏈長就等於素因子的個數，最長鏈條數就是這些素因子的排列組合數。題目資料量較大，所以先打個素數表。

程式碼清單：

#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;

const int maxp = 10000 + 5;
const int maxn = 1100000 + 5;

int sum,X;
int prime[maxp];
bool vis[maxn];
int num[maxp],k; //存每個素因子的個數

void init(){  //素數篩法打表
    sum=0;
    memset(vis,true,sizeof(vis));
    for(int i=2;i<=maxn;i++){
        if(!vis[i]) continue;
        prime[sum++]=i;
        if(i>(int)sqrt(maxn)) continue;
        for(int j=i*i;j<=maxn;j+=i)
            vis[j]=false;
    }
}

ll get_max(){  //得到最大鏈長，即素因子的個數
    ll ans=0;
    for(int i=0;i<k;i++)
        ans+=(ll)num[i];
    return ans;
}

ll get_sum(){  //得到鏈的種類數,即素因子的排列組合數
    ll fenzi=1;
    ll Max=get_max();
    for(ll i=2;i<=Max;i++)
        fenzi*=i;
    for(int i=0;i<k;i++){
        for(int j=2;j<=num[i];j++){
            fenzi/=(ll)j;
        }
    }
    return fenzi;
}

void solve(){
    k=0;
    memset(num,0,sizeof(num));
    for(int i=0;i<sum;i++){
        if(X%prime[i]==0){
            while(X%prime[i]==0){
                num[k]++;
                X/=prime[i];
            }
            k++;
        }
        if(X==1){
            break;
        }
    }
    ll ans_max=get_max();
    ll ans_sum=get_sum();
    printf("%I64d %I64d\n",ans_max,ans_sum);
}

int main(){
    init();
    while(scanf("%d",&X)!=EOF){
        solve();
    }return 0;
}