B - Link/Cut Tree

Problem description

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

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.

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解题思路：题目的意思就是给定一个区间[l,r]和一个底数为k，要求输出在区间内以k为底数的所有指数值。题目比较简单，但有个坑，就是假设当r=10^18时，k=10^9，因为r刚好在题目给定的范围内，所以接下来的t*=k即t=10^9*10^18=10^27结果远远超过unsigned long long最大值（20位数），这将导致数据溢出，因此每次必须判断r除以t得到的倍数是否不小于k，即（r/t）>=k？如果是才可进行相乘（表明相乘之后不大于r），否则break，立即退出，避免数据溢出。

AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 int main(){
 5     LL l,r,k,t=1;bool flag=false;
 6     cin>>l>>r>>k;
 7     while(t<=r){
 8         if(flag && t<=r)cout<<' '<<t;
 9         if(!flag && t>=l){cout<<t;flag=true;}//先输出第一个数
10         if((r/t)<k)break;//这一步必须加，不然可能会爆long long，即溢出
11         t*=k;
12     }
13     if(!flag)cout<<"-1"<<endl;
14     else cout<<endl;
15     return 0;
16 }