CodeForces-485C Bits(乱搞)

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.

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

题解:设L的二进制长度为len1,R的二进制长度为len2,分三种情况,①R的二进制全为1,直接输出r;②len2>len1且R的二进制不全为1,输出二进制长度为len2-1且二进制全为1的数;

③len2==len1,从高位到低位找到一个R的那一位是1,L的那一位是0的数位,再判断R在那一位之后是否全为1,如果是则把L的那一位之后且包括那一位全部变成1,如果不是则把

L的那一位之后的全部变成1,最后输出把L变成十进制再输出

#include
#include
#include
#include
#include
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define FIN freopen("in.txt","r",stdin)
using namespace std;
typedef long long LL;
typedef pair PII;
int num1[66],num2[66];
int main(){
    int n;
    LL l,r;
   // FIN;
    scanf("%d",&n);
    while(n--){
        scanf("%I64d%I64d",&l,&r);
        if(l==r) {printf("%I64d\n",r);continue;}
        memset(num1,0,sizeof(num1));
        memset(num2,0,sizeof(num2));
        int len1=0,len2=0,cnt=0;
        LL tmp=r;
        while(l>0){
            num1[len1++]=(l&1);
            l>>=1;
        }
        while(r>0){
            cnt+=(r&1);
            num2[len2++]=(r&1);
            r>>=1;
        }
        if(cnt==len2) printf("%I64d\n",tmp);
        else if(len1=0;i--)
                if(num1[i]==0&&num2[i]==1) {index=i;flag=1;break;}
            cnt=0;
            for(int i=index;i>=0;i--) if(num2[i]) cnt++;
            if(cnt==index+1) for(int i=index;i>=0;i--) num1[i]=1;
            else for(int i=index-1;i>=0;i--) num1[i]=1;
            for(int i=0;i