Claris and XOR

Problem Description

Claris loves bitwise operations very much, especially XOR, because it has many beautiful features. He gets four positive integers a,b,c,da,b,c,d that satisfies a\leq ba≤b and c\leq dc≤d. He wants to choose two integers x,yx,y that satisfies a\leq x\leq ba≤x≤b and c\leq y\leq dc≤y≤d, and maximize the value of x~XOR~yx XOR y. But he doesn't know how to do it, so please tell him the maximum value of x~XOR~yx XOR y.

Input

The first line contains an integer T\left(1\leq T\leq10,000\right)T(1≤T≤10,000)——The number of the test cases. For each test case, the only line contains four integers a,b,c,d\left(1\leq a,b,c,d\leq10^{18}\right)a,b,c,d(1≤a,b,c,d≤10​18​​). Between each two adjacent integers there is a white space separated.

Output

For each test case, the only line contains a integer that is the maximum value of x~XOR~yx XOR y.

Sample Input

2
1 2 3 4
5 7 13 15

Sample Output

Copy

6
11

Hint

In the first test case, when and only when x=2,y=4x=2,y=4, the value of x~XOR~yx XOR y is the maximum. In the second test case, when and only when x=5,y=14x=5,y=14 or x=6,y=13x=6,y=13, the value of x~XOR~yx XOR y is the maximum.

 

 

 

这游戏真难,之前写过一道类似的,就直接敲了...o(n)...超时....

 

 

TLE代码:

 #include <vector>
 #include <map>
 #include <set>
 #include <algorithm>
 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <cstdlib>
 #include <string>
 #include <cstring>
 #include <queue>
 using namespace std;
 #define INF 0x3f3f3f3f
 #define ll long long

 int const MAX = ;
 int n;

 struct Trie
 {
     int root, tot, next[MAX][], end[MAX];
     inline int node()
     {
         memset(next[tot], -, sizeof(next[tot]));
         end[tot] = ;
         return tot ++;
     }

     inline void Init()
     {
         tot = ;
         root = node();
     }

     inline void insert(ll x)
     {
         int p = root;
         for(int i = ; i >= ; i--)
         {
             int ID = (( << i) & x) ?  : ;
             if(next[p][ID] == -)
                 next[p][ID] = node();
             p = next[p][ID];
         }
         end[p] = x;
     }

     inline int search(int x)
     {
         int p = root;
         for(int i = ; i >= ; i--)
         {
             int ID = (( << i) & x) ?  : ;
             if(ID == )
                 p = next[p][] != - ? next[p][] : next[p][];
             else
                 p = next[p][] != - ? next[p][] : next[p][];
         }
         return x ^ end[p];
     }

 }trie;

 int a[],b[];
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--)
     {
         int n=;
         int WTF = ,ans=, x;
         trie.Init();
         for(int i = ; i < ; i++)
         {
             scanf("%d", &a[i]);
         }
         for(int i = ; i < ; i++)
         {
             scanf("%d", &b[i]);
         }
         for(int i=a[]; i<=a[]; i++){
             //WTF=0;
             //trie.insert(1);
             //WTF = WTFx(WTF, trie.search(1));
             for(int j=b[]; j<=b[]; j++){
                 trie.Init();
                 trie.insert(i);
                 WTF = max(WTF, trie.search(i));
                 trie.insert(j);
                 WTF = max(WTF, trie.search(j));
                 ans=max(WTF,ans);
             }
         }
         printf("%d\n", WTF);
     }
 }