codeforces C. Bits 贪心-蒲公英云

codeforces C. Bits 贪心

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 l**i, r**i — the arguments for the corresponding query (0 ≤ l**i ≤ r**i ≤ 1018).

Output

For each query print the answer in a separate line.

Examples

input

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

题意: 给定l和r, 求l和r之间的数中二进制中1的个数最多的数是哪一个, 如果有多个答案, 输出最小的那个数.

分析: 我们可以贪心的一步一步的将l的最低位0 变成1, 因为要使1的个数最多, 直到l再取大于r的时候就不能取了,这时候l的二进制中1的数目最多, 并且是最小的.

#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=100010,MOD=1e9+7;
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        ll l,r;
        cin>>l>>r;
        ll ans=l;
        while(l<=r)
        {
            ans = l;
            for(int i=0;i<61;i++){
                if(!(l&(1LL<<i))){
                    l |= 1LL<<i;
                    break;
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}
#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=100010,MOD=1e9+7;
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        ll l,r;
        cin>>l>>r;
        ll ans=l;
        while(l<=r)
        {
            ans = l;
            ll m = (~l)&(-(~l));
            l += m;
        }
        cout<<ans<<endl;
    }
    return 0;
}