【分块打表】Gym - 100923K - Por Costel and the Firecracker

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

因为卡内存，所以不能把答案的表全打出来，但是可以每隔100记录一次答案，这样只需要开10w的数组。然后每次询问的时候，从最近的记录的答案开始暴力，不超过100次就能得到答案。

#include<cstdio>
using namespace std;
#define MOD 10000003
typedef long long ll;
int n,a,b,x1,q,q1;
int anss[100010];
int main()
{
	freopen("pocnitoare.in","r",stdin);
	freopen("pocnitoare.out","w",stdout);
//	freopen("k.in","r",stdin);
	scanf("%d%d%d%d%d%d",&n,&a,&b,&x1,&q,&q1);
	int now=x1;
	anss[1]=now;
	for(int i=2;i<=10000003;++i)
	  {
	  	now=(int)((((ll)now*(ll)(i-1)%(ll)n)%(ll)n+(ll)a%(ll)n)%(ll)n);
	  	if(i%100==1)
	  	  anss[i/100+1]=now;
	  }
//	int now=anss[q1/100+1];
//	int tmp=q1%100-1;
//	for(int i=1;i<=tmp;++i)
//	  now=(int)((((ll)now*(ll)(i-1)%(ll)n)%(ll)n+(ll)a%(ll)n)%(ll)n);
//	printf("%d\n",now);
	for(int i=1;i<=q;++i)
	  {
	  	if(i!=1)
	  	  q1=((int)((ll)(i-1)*(ll)now%(ll)MOD)+b%MOD)%MOD+1;
	  	now=anss[(q1-1)/100+1];
		for(int j=(q1-1)/100*100+2;j<=q1;++j)
	  	  now=(int)((((ll)now*(ll)(j-1)%(ll)n)%(ll)n+(ll)a%(ll)n)%(ll)n);
		printf("%d\n",now);
	  }
	return 0;
}