【Codeforces Gym 100187A】

A - A Gym - 100187A

The world famous scientist Innokentiy has just synthesized the potion of immortality. Unfortunately, he put the flask with this potion on the shelf where most dangerous poisons of all time were kept. Now there are n flasks on this shelf, and the scientist has no idea which of them contains the potion of immortality.

Fortunately, Innokentiy has an infinite amount of rabbits. But the scientist doesn’t know how exactly these potions affect rabbits. There is the only thing he knows for sure. If rabbit tastes exactly k potions from the flasks on the shelf, it will survive if there was the immortality potion among them and die otherwise. If rabbit tastes the number of potions different from k, the result will be absolutely unpredictable, so the scientist won’t make such experiments no matter what.

The scientist intends to minimize the amount of lost rabbits while searching for the potion of immortality. You should determine how many rabbits he has to sacrifice in the worst case.

Input

The only line contains two integers separated by space — n and k (1 ≤ n ≤ 2000, 1 ≤ k ≤ n) — the number of flasks on the Innokentiy’s shelf and the number of potions Innokentiy will give to a single rabbit to taste.

Output

If the scientist cannot determine which flusk contains the potion of immortality, output «-1». Otherwise, output a single integer — the minimal number of rabbits that Innokentiy will sacrifice in the worst case while searching for the potion of immortality.

Example

Input

3 2

Output

1

Input

4 2

Output

2

喝解药的兔子一直不会死～～

AC代码：

#include<cstdio>
int main()
{
int N,K;
scanf("%d %d",&N,&K);
if(N == 1) printf("0\n");
else if(N == K) printf("-1\n");
else if(K == 1) printf("%d\n",N - 1);
else{
if(N % K == 0 || N % K == 1) printf("%d\n", N / K);
else printf("%d\n",N / K + 1);
}
return 0;
}