Table Tennis Game 2

Description

Misha and Vanya have played several table tennis sets. Each set consists of several serves, each serve is won by one of the players, he receives one point and the loser receives nothing. Once one of the players scores exactly k points, the score is reset and a new set begins.

Across all the sets Misha scored a points in total, and Vanya scored b points. Given this information, determine the maximum number of sets they could have played, or that the situation is impossible.

Note that the game consisted of several complete sets.

Input

The first line contains three space-separated integers k, a and b (1 ≤ k ≤ 109, 0 ≤ a, b ≤ 109, a + b > 0).

Output

If the situation is impossible, print a single number -1. Otherwise, print the maximum possible number of sets.

Sample Input

Input

11 11 5

Output

1

Input

11 2 3

Output

-1

Sample Output

Hint

Note that the rules of the game in this problem differ from the real table tennis game, for example, the rule of "balance" (the winning player has to be at least two points ahead to win a set) has no power within the present problem.

题意：两个人进行乒乓球比赛，比赛分为若干局，每一局分为若干回合，在每一回合中，胜利的人的一分，输的人不得分，若其中有一个人在一局中先得到k分，则这一局结束，两人分数都重置，给你三个数，k,a,b，k代表上面的k，a，b分别代表两人最后的总得分（每一局得分之和），问你两个人最多进行了多少回合比赛，如果不存在这种状况，则输出-1. 
解题思路：首先我们要知道，在一定的分数下，如果要得到更多的局数，肯定先让每一局为一个人的k分，一个人的0分，然后剩下的分数补起来就行，特别需要注意的是，最后结束肯定是以一整局结束为准，比如为什么样例2中11，2，3是输出-1,因为两人的得分分别是2，3都小于11，如果最后游戏结束的话，肯定有一个人得到11分，所以这种情况不会出现，所以我们需要特别注意的是，当两个人的分数都小于k时，一定是无解，而两个人的分数都大于k时，答案时a/k + b/k,最复杂的情况是其中一个人的分数小于k,另外一个人的分数大于k,这种情况需要特判一下，那个大于k 的那个人的分数是不是k的整数倍。如果不是，则无解。

#include <stdio.h>
int main()
{
    long long k,a,b;
    while(~scanf("%lld%lld%lld",&k,&a,&b))
    {
        if(a<k&&b<k)
            printf("-1
");
        else
        if(a<k&&b>=k)
        {
            if(b%k!=0)
                printf("-1
");
            else
                printf("%d
",b/k);
        }
        else
        if(b<k&&a>=k)
        {
            if(a%k!=0)
                printf("-1
");
            else
                printf("%d
",a/k);
        }
        else
            printf("%d
",a/k+b/k);
    }
    return 0;
}