Dreamoon and Stairs

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---

title: Dreamoon and Stairs
date: 2018-08-10 20:13:08
tags:

• Codeforces
• ACM
• 数学
  categories:
• 数学

---

题目链接

Dreamoon wants to climb up a stair of n steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer m.

What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition?

Input

The single line contains two space separated integers n, m (0 < n ≤ 10000, 1 < m ≤ 10).

Output

Print a single integer — the minimal number of moves being a multiple of m. If there is no way he can climb satisfying condition print  - 1 instead.

Examples

input

10 2

output

6

input

3 5

output

-1

Note

For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5.

有一个n级台阶，每次可以走一级或两级，问最少的步数是多少，且步数必须是m的倍数。
找一下数学公式就好了。
具体看代码。

#include<bits/stdc++.h>
using namespace std;

int main(){
    int x,n,m;
    cin>>n>>m;
    if(n<m){
        cout<<-1<<endl;
        return 0;
    }
    if(n==m){
        cout<<n<<endl;
        return 0;
    }
    if(n%2==0){
        x=n/2%m;
        if(x==0) cout<<n/2<<endl;
        else cout<<n/2+m-x<<endl;
    }else if(n%2!=0){
        x=(n/2+1)%m;
        if(x==0) cout<<n/2+1<<endl;
        else cout<<(n/2+1)+m-x<<endl;
    }
	return 0;
}