Description

Most of the time when rounding a given number, it is customary to round to some multiple of a power of 10. However, there is no reason why we cannot use another multiple to do our rounding to. For example, you could round to the nearest multiple of 7, or the nearest multiple of 3.
    Given an int n and an int b, round n to the nearest value which is a multiple of b. If n is exactly halfway between two multiples of b, return the larger value.

Input

Each line has two numbers n and b,1<=n<=1000000,2<=b<=500

Output

The answer,a number per line.

Sample Input

5 10

4 10

Sample Output

10

0

Hint

#include<stdio.h>

int main()

{

    int n,b;

    while(scanf("%d%d",&n,&b)!=EOF)

    {

        int i=1;

        while(n>b*i) i++;

        int temp1=b*i;

        int temp2=b*(i-1);

        if(temp1-n>n-temp2)

            printf("%d\n",temp2);

        else

            printf("%d\n",temp1);

    }

    return 0;

}

/**********************************************************************

	Problem: 1040

	User: song_hai_lei

	Language: C++

	Result: AC

	Time:0 ms

	Memory:1120 kb

**********************************************************************/