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 thex, such thatl ≤ 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 integersl_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.

Example

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

结果要的是最多1的数，相同数量的1的数中，输出，最小值。我们知道，从低位不断向上添加，这样的话，就是最优的解。因为我们中间不会放置0。

所以我们只需要不断地在left的二进制数中，不断地从低位向高位添加1，一直到不能添加，这样得到的结果就是最优解。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;typedef long long ll;int main(){    int n;    scanf("%d",&n);    for(int i=1;i<=n;i++)    {        ll left,right;        scanf("%I64d %I64d",&left,&right);        ll val=1;        ll ans=left;        while(left<right)        {            left = left | val;//不断地往高位添加1            if(left>right)                break;            ans = left;            val = val<< 1;        }        printf("%I64d\n",ans);    }    return 0;}