Codeforces Round #629 (Div. 3) B. K-th Beautiful String（找规律）

For the given integer nn (n>2n>2 ) let's write down all the strings of length nn which contain n−2n−2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.

Recall that the string ss of length nn is lexicographically less than string tt of length nn , if there exists such ii (1≤i≤n1≤i≤n ), that si<tisi<ti , and for any jj (1≤j<i1≤j<i ) sj=tjsj=tj . The lexicographic comparison of strings is implemented by the operator < in modern programming languages.

For example, if n=5n=5 the strings are (the order does matter):

1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa

It is easy to show that such a list of strings will contain exactly n⋅(n−1)2n⋅(n−1)2 strings.

You are given nn (n>2n>2 ) and kk (1≤k≤n⋅(n−1)21≤k≤n⋅(n−1)2 ). Print the kk -th string from the list.

Input

The input contains one or more test cases.

The first line contains one integer tt (1≤t≤1041≤t≤104 ) — the number of test cases in the test. Then tt test cases follow.

Each test case is written on the the separate line containing two integers nn and kk (3≤n≤105,1≤k≤min(2⋅109,n⋅(n−1)2)3≤n≤105,1≤k≤min(2⋅109,n⋅(n−1)2) .

The sum of values nn over all test cases in the test doesn't exceed 105105 .

Output

For each test case print the kk -th string from the list of all described above strings of length nn . Strings in the list are sorted lexicographically (alphabetically).

Example

Input

Copy

7
5 1
5 2
5 8
5 10
3 1
3 2
20 100

Output

Copy

aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa
因为b只有两个，所以我观察了一下b的出现的位置，发现很有规律。假设最后一位是1，第一位是n，则第一个b出现的位置类似1 22 333 4444......只不过要再偏移一位；第二个出现的b的位置为 1 12 123 1234......这样只需要判断一下k在哪一组里，把相应的位置打上标记即可。由于所有test case的n的和不超过1e5，所以暴力乱搞即可。

#include <bits/stdc++.h>
using namespace std;
bool b[100005]={0};
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        long long n,k;
        memset(b,0,sizeof(b));
        cin>>n>>k;//所有n的和不超过1e5 
        long long i;
        long long nxt;
        for(i=1;i<=n;i++)
        {
            if(i*(i-1)/2<k&&i*(i+1)/2>=k)
            {
                nxt=i+1;
                break;
            }
        }
        b[n-nxt+1]=1;
        b[n-(k-(nxt-1)*(nxt-2)/2)+1]=1;
        for(i=1;i<=n;i++)
        {
            if(b[i]==0)printf("a");
            else printf("b");
        }
        cout<<endl;
    }
    return 0;
}
//10
//5 1
//5 2
//5 3
//5 4
//5 5
//5 6
//5 7
//5 8
//5 9
//5 10