CodeForces 485C Bits[贪心 二进制]

C. Bits

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Let's denote as  the number of bits set ('1' bits) in the binary representation of the non-negative integer x.

You are given multiple queries consisting of pairs of integers l and r. For each query, find the x, such that l ≤ x ≤ r, and  is maximum possible. If there are multiple such numbers find the smallest of them.

Input

The first line contains integer n — the number of queries (1 ≤ n ≤ 10000).

Each of the following n lines contain two integers li, ri — the arguments for the corresponding query (0 ≤ li ≤ ri ≤ 1018).

Output

For each query print the answer in a separate line.

Examples

input

3

1 2

2 4

1 10

output

1

3

7

Note

The binary representations of numbers from 1 to 10 are listed below:

110 = 12

210 = 102

310 = 112

410 = 1002

510 = 1012

610 = 1102

710 = 1112

810 = 10002

910 = 10012

1010 = 10102

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题意：求l和r之间二进制1的个数最多的数字x

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首先证明x一定在l的基础上

然后从低位贪心把0变成1

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;
typedef long long ll;
inline ll read(){
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
ll T,l,r;
int main(int argc, const char * argv[]) {
    T=read();
    while(T--){
        l=read();r=read();
        ll t=1;
        while(l<r){
            if((l|t)>r) break;
            l|=t;
            t<<=1;
        }
        printf("%I64d
",l);
    }
    return 0;
}