https://odzkskevi.qnssl.com/76ee6a37a5e7b6e0a1da0f46372ab4da?v=1512194397
题意：给你一个n，分解这个n，然后得到他的因子乘积，并且前后因子的差距最多为1。
比如可以 2 3 。 2 2。 但不能2 3 4。
做法：首先考虑如何得到这个n的因子x，一定是要靠近这个x也就是 x-1 x x+1，这三种与x接近的才能构成答案。
一直觉得是根号n的时间处理这个问题，但是n是1e18，后来看了别人的代码，用了二分的方法进行找因子，
具体是枚举幂项，二分1-n，找到mid^j是n的因子，腻害腻害，确实不会这种操作。最后就是要么全是mid^j构成n，
或者mid^j*(mid+1)^k构成n，记录答案即可。
/// .-~~~~~~~~~-._ _.-~~~~~~~~~-.
/// __.' ~. .~ `.__
/// .'// \./ \\`.
/// .'// | \\`.
/// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`.
/// .'//.-" `-. | .-' "-.\\`.
/// .'//______.============-.. \ | / ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma GCC optimize(2)
#pragma comment(linker, "/STACK:102400000,102400000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;
#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=~0U>>1;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e5+10;
const int maxx=1e6+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}
inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}
void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;
//void readString(string &s)
//{
// static char str[maxn];
// scanf("%s", str);
// s = str;
//}
LL q[100][100];
int a[100],cnt=0;
LL F(LL a,LL i)
{
LL l=1,r=a,b;
W(l<r)
{
LL mid=(l+r+1)/2;
b=a;
FOR(1,i,j) b/=mid;
if(b) l=mid;
else r=mid-1;
}
return l;
}
LL n;
void solve()
{
S_1(n);
if(n==1) {puts("-1");return ;}
int i;
for(i=60;i;i--) if(n>>i) break;
FOR(1,i,j) if(n==1ll<<j) {puts("-1");return ;}
FOR(1,i,j)
{
LL root=F(n,j);
LL xx=n;
int tot1=0,tot2=0;
while(xx%root==0)
{
xx/=root;
tot1++;
}
while(xx%(root+1)==0)
{
xx/=(root+1);
tot2++;
}
if(xx==1&&tot1+tot2==j)
{
a[++cnt]=j;
for(int k=1;k<=tot1;k++)
q[cnt][k]=root;
for(int k=tot1+1;k<=j;k++)
q[cnt][k]=root+1;
}
}
printf("%d\n",cnt);
for(int i=1;i<=cnt;i++)
{
printf("%d",a[i]);
for(int j=1;j<=a[i];j++)
{
printf(" %lld",q[i][j]);
}
puts("");
}
}
int main()
{
freopen( "little.in " , "r" , stdin );
freopen( "little.out" , "w" , stdout );
int t=1;
//init();
//s_1(t);
for(int cas=1;cas<=t;cas++)
{
//printf("Case #%d: ",cas);
solve();
}
}