A little girl loves problems on bitwise operations very much. Here’s one of them.
You are given two integers l and r. Let’s consider the values of for all pairs of integers a and b (l ≤ a ≤ b ≤ r). Your task is to find the maximum value among all considered ones.
Expression means applying bitwise excluding or operation to integers x and y. The given operation exists in all modern programming languages, for example, in languages C++ and Java it is represented as “^”, in Pascal — as «xor».
Input
The single line contains space-separated integers l and r (1 ≤ l ≤ r ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print a single integer — the maximum value of for all pairs of integers a, b (l ≤ a ≤ b ≤ r).
Example
Input
1 2
Output
3
Input
8 16
Output
31
Input
1 1
Output
0

【題解】
首先，異或值要為1，那麼兩個數的對應二進位制位要不同，而異或值要最大，則二進位制的高位要儘可能的為1，所以這就是切入點，從給定的區間上下限入手（l和R），從l和r二進位制的最高位開始比較，如果出現對應位異或值位1，就從該處開始，低位都置為1，此時其表示的數就是最大異或值了。

#include<stdio.h>
int  main()
 {
 long long l,r,t,w;
 while(scanf("%lld%lld",&l,&r)!=EOF)
 {  
   if(r==l)printf("0\n");
   else
   {w=0;
       while
 
(r!=l&&r!=0)//相等或最大值為0 結束                101011 
       {                //相等則證明 此位之前的異或不可能為1  101110 
          w++;
          l=l>>1;//從右向左移動 
          r=r>>1;
       }
        t=1;
       while(w--)t=t*2;
       printf("%lld\n",t-1); 
   }
 }
   return 0;
 }