Codeforces Round #276 (Div. 1) A. Bits Greedy

A. Bits

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- l ≤ x ≤ R, and is maximum Possibl E. If there is 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 II integers li, ri -t He arguments for the corresponding query (0≤ li ≤ ri ≤101 8).

Output

For each query, print the answer in a separate line.

Sample Test (s) input

3
1 2
2 4
1 10

Output

1
3
7

Note

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

1= 12

2=2

3=2

4=2

5= 1012

6=2

7= 1112

8=2

9= 10012

Ten= 10102

Test Instructions : To n inquiries, each time you ask the number between l,r, in the binary under the 1-digit number of which is the most

Solution: We build from the L, in 0-bit, greedy from small to large-bit, must be the best

///1085422276#include<bits/stdc++.h>using namespacestd;#pragmaComment (linker, "/stack:102400000,102400000")using namespacestd; typedefLong Longll;typedef unsignedLong Longull;#defineMem (a) memset (A,0,sizeof (a))#definePB Push_backinline ll read () {ll x=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9'){        if(ch=='-') f=-1; ch=GetChar (); }     while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar (); }returnx*F;}//****************************************Const intn=6000+ -;#defineMoD 10000007#defineINF 1000000001#defineMAXN 10000intd[ the];ll Test (ll X,ll y) { for(intI=0;i< the; i++) {        if(! (x& (1ll<<i)) &&x+ (1ll<<i) <=y) x+ = (1ll<<i); }    returnx;}intMain () {intn=read (); ll L,r;  for(intI=1; i<=n;i++) {cin>>l>>R; cout<<test (l,r) <<Endl; }    return 0;}

Code

Codeforces Round #276 (Div. 1) A. Bits Greedy