Codeforces Round #276 (Div. 2)(C)

C. Bits

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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.

Sample test(s)

input

31 22 41 10

output

137

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

#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>using namespace std;__int64  bit[65];int main() {   bit[0]=1;    for(int i = 1; i <= 63; i++) {        bit[i] = 2*bit[i-1];    }    int n;    scanf("%d",&n);    while(n--) {         __int64 l,r;         scanf("%I64d%I64d",&l,&r);         int j;         __int64 sum = 0;         for(j = 0; sum <=r; j++)         {             sum += bit[j];         }         j--;         while(sum > r)         {             sum -= bit[j];             if(sum < l) sum += bit[j];             j--;         }        printf("%I64d\n",sum);    }    return 0;}