Codeforces Round #456 (Div. 2) B. New Year's Eve-优快云博客

本文介绍了一个关于新年夜糖果选择的问题，目标是找到从n种不同口味糖果中选取k个，使得这些糖果口味异或和最大。文章给出了两种解决思路及代码实现。

Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains n sweet candies from the good ol' bakery, each labeled from 1 to n corresponding to its tastiness. No two candies have the same tastiness.

The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!

A xor-sum of a sequence of integers a₁, a₂, ..., a_m is defined as the bitwise XOR of all its elements: $\text{[math]}$ , here $\text{[math]}$ denotes the bitwise XOR operation; more about bitwise XOR can be found here.

Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than k candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.

Input

The sole string contains two integers n and k (1 ≤ k ≤ n ≤ 10¹⁸).

Output

Output one number — the largest possible xor-sum.

Examples

Input

4 3

Output

Input

6 6

Output

Note

In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.

In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7.

一眼题，当k为1输出本身，k不为1时，要让异或最大，则最高位以下（包括最高位）的的二进制全为1，或者像我第一反应一样打表2333(蠢哭)...

正常代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    ll n,k;
    scanf("%lld%lld",&n,&k);
    if(k==1)printf("%lld\n",n);
    else
    {
        ll r=0;
        while(n)n>>=1,r=r<<1|1;
        printf("%lld\n",r);
    }
    return 0;
}

蠢哭代码：

#include<bits/stdc++.h>
//******************************************************
#define lrt (rt*2)
#define rrt  (rt*2+1)
#define LL long long
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
#define exp 1e-8
//***************************************************
#define eps             1e-8
#define inf             0x3f3f3f3f
#define INF             2e18
#define LL              long long
#define ULL             unsigned long long
#define PI              acos(-1.0)
#define pb              push_back
#define mk              make_pair

#define all(a)          a.begin(),a.end()
#define rall(a)         a.rbegin(),a.rend()
#define SQR(a)          ((a)*(a))
#define Unique(a)       sort(all(a)),a.erase(unique(all(a)),a.end())
#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(min(a,b),min(c,d))
#define max4(a,b,c,d)   max(max(a,b),max(c,d))
#define max5(a,b,c,d,e) max(max3(a,b,c),max(d,e))
#define min5(a,b,c,d,e) min(min3(a,b,c),min(d,e))
#define Iterator(a)     __typeof__(a.begin())
#define rIterator(a)    __typeof__(a.rbegin())
#define FastRead        ios_base::sync_with_stdio(0);cin.tie(0)
#define CasePrint       pc('C'); pc('a'); pc('s'); pc('e'); pc(' '); write(qq++,false); pc(':'); pc(' ')
#define vi              vector <int>
#define vL              vector <LL>
#define For(I,A,B)      for(int I = (A); I < (B); ++I)
#define FOR(I,A,B)      for(int I = (A); I <= (B); ++I)
#define rFor(I,A,B)     for(int I = (A); I >= (B); --I)
#define Rep(I,N)        For(I,0,N)
#define REP(I,N)        FOR(I,1,N)
using namespace std;
const int maxn=1e5+10;
vector<int>Q[maxn];
const int MOD=1e9+7;
LL lev[65],sta[65];
void pre()
{
    Rep(i,62) lev[i]=(1LL<<i)-1;
    sta[0]=1;
    REP(i,61) sta[i]=sta[i-1]<<1;
    Rep(i,62) lev[62]+=sta[i];
}

int main()
{
    FastRead;
    LL k,n;
    pre();
    cin>>n>>k;
        if(k==1) cout<<n<<endl;
        else
        {
            for(int i=0;i<=61;i++)
                if(n>=lev[i]&&n<=lev[i+1])
                {
                    cout<<lev[i+1]<<endl;
                    break;
                }
        }

    return 0;
}