【找规律】Gym - 100923L - Por Costel and the Semipalindromes

semipal.in / semipal.out

Por Costel the pig, our programmer in-training, has recently returned from the Petrozaporksk training camp. There, he learned a lot of things: how to boil a cob, how to scratch his belly using his keyboard, etc... He almost remembers a programming problem too:

A semipalindrome is a word  for which there exists a subword  such that  is a prefix of  and  (reverse ) is a suffix of . For example, 'ababba' is a semipalindrom because the subword 'ab' is prefix of 'ababba' and 'ba' is suffix of 'ababba'.

Let's consider only semipalindromes that contain letters 'a' and 'b'. You have to find the -th lexicographical semipalindrome of length .

Por Costel doesn't remember if the statement was exactly like this at Petrozaporksk, but he finds this problem interesting enough and needs your help to solve it.

Input

On the first line of the file semipal.in, there is an integer  () representing the number of test cases. On the next  lines there are 2 numbers,  ( and K  where  is the number of semipalindromes of length .

Output

In the output file semipal.out, there should be  lines, the -th of which should contain the answer for the -th test.

Example

Input

2
5 1
5 14

Output

aaaaa
bbabb

 

显然只需要保证开头和结尾字母相同，就一定是合法的啦。

然后就把中间的部分得到即可。

#include<cstdio>
using namespace std;
typedef long long ll;
int T,n;
ll m;
char a[70];
int main()
{
//	freopen("l.in","r",stdin);
	freopen("semipal.in","r",stdin);
	freopen("semipal.out","w",stdout);
	scanf("%d",&T);
	for(;T;--T)
	  {
	  	scanf("%d%I64d",&n,&m);
//	  	ll all=1ll<<(n-1);
	  	bool flag=0;
	  	if(m>(1ll<<(n-2)))
	  	  {
	  	  	m-=(1ll<<(n-2));
	  	  	flag=1;
	  	  }
	  	--m;
	  	for(int i=1;i<=n-2;++i)
	  	  {
	  	  	a[i]=(m%2 ? 'b' : 'a');
	  	  	m/=2;
	  	  }
	  	if(flag)
	  	  {
	  	  	putchar('b');
	  	  	for(int i=n-2;i>=1;--i)
	  	  	  putchar(a[i]);
	  	  	puts("b");
	  	  }
	  	else
	  	  {
	  	  	putchar('a');
	  	  	for(int i=n-2;i>=1;--i)
	  	  	  putchar(a[i]);
	  	  	puts("a");
	  	  }
	  }
	return 0;
}