Codeforces round #256 (Div. 2) E Divisors

E. Divisors

Bizon the champion isn't just friendly, he also is a rigorous coder.

Let's define functionF(A), WhereAIs a sequence of integers. FunctionF(A) Returns the following sequence: First all divisorsA_{1 go in the increasing order, then all divisorsA2 go in the increasing order, and so on till the last element of SequenceA. For example,F([2, 9, 1_1]) records = values [1, 1_2, 1_1, 1_3, 1_9, 1_1].}

Let's determine the sequenceX_{I, For integerI(ILimit ≥ limit 0 ):X0 bytes = average [X] ([X] Is a sequence consisting of a single numberX),XISignature = SignatureF(XIHistogram-interval 1 )(ILatency> limit 0). For example,XKeys = keys 6 we getX0 bytes = bytes [6],X1 rows = Second [1, second 2, second 3, second 6],X2 records = records [1, records 1, records 2, records 1, records 3, records 1, records 2, records 3, records 6].}

Given the numbersXAndK, Find the sequenceX_{K. As the answer can be rather large, find only the first 10^{5 elements of this sequence.}}

Input

A single line contains two space-separated integers-X(1 digit ≤ DigitXLimit ≤ limit 10^{12) andK(0 bytes ≤ bytesKLimit ≤ limit 1018 ).}

Output

Print the elements of the sequenceX_{KIn a single line, separated by a space. If the number of elements exceeds 10^{5, then print only the first 105 elements.}}

Sample test (s) Input

6 1

Output

1 2 3 6 

Input

4 2

Output

1 1 2 1 2 4 

Input

10 3

Output

1 1 1 2 1 1 5 1 1 2 1 5 1 2 5 10 

 1 #include<stdio.h> 2 #include<map> 3 #include<algorithm> 4 #define MAXN 20000000 5 using namespace std; 6 typedef long long LL; 7 LL ys[9000];int tot=0;int tt2=0; 8 int tail[9000];int head[9000]; 9 int aft[MAXN];LL p[MAXN];10 LL n,k;11 map<LL,int>bh;12 void line(int j,int i)13 {14      tt2++;if(!tail[j])head[j]=tt2;p[tt2]=i;aft[tail[j]]=tt2;tail[j]=tt2;15 }16 void init()17 {18      for(LL i=1;i*i<=n;i++)19      if(n%i==0)20      {21                ys[++tot]=i;22                if(i*i!=n)23                ys[++tot]=n/i;24                }25      sort(ys+1,ys+1+tot);26      for(int i=1;i<=tot;i++)bh[ys[i]]=i;27      for(int i=1;i<=tot;i++)28      for(int j=1;j<=i;j++)29      if(ys[i]%ys[j]==0)line(i,j);30 }31 int dfs(int now,LL dep,int need)32 {33     if(ys[now]==1)34     {35                   printf("1 ");36                   return 1;37                   }38      if(dep==k)39      {40                int us=need;41                for(int u=head[now];u&&need;need--,u=aft[u])42                printf("%I64d ",ys[p[u]]);43                return us-need;44                }45      int us=need;46      for(int u=head[now];u&&need;need-=dfs(p[u],dep+1,need),u=aft[u]);47      return us-need;48 }49 int main()50 {51     scanf("%I64d%I64d",&n,&k);if(k>100000)k=100000;52     init();53     if(!k)54     {55           printf("%I64d\n",n);56           return 0;57           }58     dfs(bh[n],1,100000);59     return 0;60 }

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