Can the greatest common divisor and bitwise operations have anything in common? It is time to answer this question.

Suppose you are given a positive integer?aa. You want to choose some integer?bb?from?11to?a?1a?1?inclusive in such a way that the?greatest common divisor (GCD)?of integers?a⊕ba⊕b?and?a&ba&b?is as large as possible. In other words, you'd like to compute the following function:

?

f(a)=max0<b<agcd(a⊕b,a&b).f(a)=max0<b<agcd(a⊕b,a&b).

Here?⊕⊕?denotes the?bitwise XOR operation, and?&&?denotes the?bitwise AND operation.

The greatest common divisor of two integers?xx?and?yy?is the largest integer?gg?such that both?xx?and?yy?are divided by?gg?without remainder.

You are given?qq?integers?a1,a2,…,aqa1,a2,…,aq. For each of these integers compute the largest possible value of the greatest common divisor (when?bb?is chosen optimally).

Input

The first line contains an integer?qq?(1≤q≤1031≤q≤103)?— the number of integers you need to compute the answer for.

After that?qq?integers are given, one per line:?a1,a2,…,aqa1,a2,…,aq?(2≤ai≤225?12≤ai≤225?1)?— the integers you need to compute the answer for.

Output

For each integer, print the answer in the same order as the integers are given in input.

Example

Input

3 2 3 5

Output

3 1 7

Note

For the first integer the optimal choice is?b=1, then a⊕b=3,a&b=0, and the greatest common divisor of?33?and?00?is?33.

For the second integer one optimal choice isb=2, then?a⊕b=1,?a&b=2, and the greatest common divisor of?1and?2?is?1.

For the third integer the optimal choice is?b=2, then?a⊕b=7,?a&b=0a&b=0, and the greatest common divisor of?7?and?0?is?7.

代碼：

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> #include<vector> #include<stack> #include<set> #include<map>using namespace std;int main() {int n;cin>>n;long long x;for(int t=0;t<n;t++){scanf("%lld",&x);long long sum=1;if(x==3){printf("1\n");}else if(x==7){printf("1\n");}else if(x==15){printf("5\n");}else if(x==31){printf("1\n");}else if(x==63){printf("21\n");}else if(x==127){printf("1\n");}else if(x==255){printf("85\n");}else if(x==511){printf("73\n");}else if(x==1023){printf("341\n");}else if(x==2047){printf("89\n");}else if(x==4095){printf("1365\n");}else if(x==8191){printf("1\n");}else if(x==16383){printf("5461\n");}else if(x==32767){printf("4681\n");}else if(x==65535){printf("21845\n");}else if(x==131071){printf("1\n");}else if(x==262143){printf("87381\n");}else if(x==524287){printf("1\n");}else if(x==1048575){printf("349525\n");}else if(x==2097151){printf("299593\n");}else if(x==4194303){printf("1398101\n");}else if(x==8388607){printf("178481\n");}else if(x==16777215){printf("5592405\n");}else if(x==33554431){printf("1082401\n");}else{for(int t=1;t<=26;t++){sum*=2;if(sum<=x&&sum*2>x){sum*=2;break;}}cout<<sum-1<<endl;}}return 0; }

?

轉載于:https://www.cnblogs.com/Staceyacm/p/10781807.html

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