codeforces-614【A精度】

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A. Link/Cut Tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the expose procedure.

Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)

Given integers l, r and k, you need to print all powers of number k within range from l to r inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!

Input

The first line of the input contains three space-separated integers l, r and k (1 ≤ l ≤ r ≤ 1018, 2 ≤ k ≤ 109).

Output

Print all powers of number k, that lie within range from l to r in the increasing order. If there are no such numbers, print "-1" (without the quotes).

Examples

input

1 10 2

output

1 2 4 8 

input

2 4 5

output

-1

Note

Note to the first sample: numbers 20 = 1, 21 = 2, 22 = 4, 23 = 8 lie within the specified range. The number 24 = 16 is greater then 10, thus it shouldn't be printed.

大意：给出一个区间，让你打印出 k 的n次方存在此区间内的数。

#include<cstdio>#include<cmath>#define LL long longusing namespace std;LL l,r,k;/* LL qpow(LL x,LL y){LL ans=1;while(y){if(y&1)ans*=x;x*=x;y>>=1;}return ans;}int main(){while(~scanf("%lld%lld%lld",&l,&r,&k)){LL a=1LL*log(l*1.0)/log(k*1.0)+1e-10; // 加精度逼近 LL b=1LL*log(r*1.0)/log(k*1.0)+1e-10;    while(qpow(k,a)<l)    {          a++;      }      while(qpow(k,b)>r)     {          b--;      }      if(a>b)      {          puts("-1");          continue;     }    LL ans=qpow(k,a);    printf("%lld ",ans);for(LL i=a+1;i<=b;i++){ans*=k;printf("%lld ",ans);}puts("");}return 0; }*/int main(){while(~scanf("%lld%lld%lld",&l,&r,&k)){bool flag=1;for(LL ans=1;;ans*=k){if(ans>=l&&ans<=r){if(!flag)printf(" ");printf("%lld",ans);flag=0;}if(ans>r/k) // 这里要提前判断，防止下一个 ans 爆 LL 而造成判断失误 break;}if(flag)puts("-1");elseputs("");}return 0;}