Codeforces Round #339 (Div. 2) A

最新推荐文章于 2020-02-12 18:28:36 发布

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最新推荐文章于 2020-02-12 18:28:36 发布

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本文链接：https://blog.youkuaiyun.com/sdut_jk17_zhangming/article/details/79562106

本文介绍了一个程序问题，即找出指定范围内所有给定基数的幂次。输入包含三个整数：范围的下限、上限及基数。程序将输出所有位于该范围内的幂次，按升序排列；若无符合条件的幂次，则输出-1。

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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).

Example

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.

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int main()
{
    long long l,r,k,sum = 1,flag = 0;
    scanf("%lld%lld%lld",&l,&r,&k);
    while(1)
    {
        if(sum>=l)
        {
            if(flag)
                printf(" ");
            printf("%lld",sum);
            flag =1;
        }
        if(sum > r/k) break;
        sum *= k;
    }
    if(!flag)
        printf("-1");
    return 0;
}