Dreamoon and Stairs上台阶

A. Dreamoon and Stairs

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 steps 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  - 1instead.

Sample test(s)

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.

题目意思：http://codeforces.com/contest/476/problem/A

n阶台阶，每次只能走2阶或则1阶，问，最少的移动次数，满足m的倍数。否则，输出-1。

想法：贪心，最少台阶嘛，每次移动全部都是2，然后逐步把2变成2个1，直到满足是m的倍数，或则全部都是1。

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<vector>#include<queue>#include<map>#include<stack>#define rt return#define bk break#define ct continue#define sf scanf#define pf printf#define ms memset#define si(n) sf("%d",&n)#define pi(n) pf("%d\n",n)#define REP0(i,n) for(int i=0;i<(n);i++)#define REP1(i,n) for(int i=1;i<=(n);i++)#define REP(i,s,n) for(int i=s;i<=(n);i++)#define db double#define op operator#define pb push_back#define LL long long#define INF 0x3fffffff#define eps 1e-8#define PI acos(-1)#define maxn 1010using namespace std;int main(){    #ifdef ACBang//    freopen("in.txt","r",stdin);    #endif    int n,m;    while(~sf("%d%d",&n,&m)){        int cnt2=n/2;        int cnt1=n%2;        while((cnt2+cnt1)%m!=0&&cnt2!=0){            cnt2--;            cnt1+=2;        }//        cout<<cnt1<<" "<<cnt2<<endl;        if((cnt1+cnt2)%m!=0)puts("-1");        else pf("%d\n",cnt1+cnt2);    }    rt 0;}