Codeforces Round #449 (Div. 2) C. DFS

C. Nephren gives a riddle

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

What are you doing at the end of the world? Are you busy? Will you save us?

Nephren is playing a game with little leprechauns.

She gives them an infinite array of strings, f0... ∞.

f0 is "What are you doing at the end of the world? Are you busy? Will you save us?".

She wants to let more people know about it, so she defines fi =  "What are you doing while sending "fi - 1"? Are you busy? Will you send "fi - 1"?" for all i ≥ 1.

For example, f1 is

"What are you doing while sending "What are you doing at the end of the world? Are you busy? Will you save us?"? Are you busy? Will you send "What are you doing at the end of the world? Are you busy? Will you save us?"?". Note that the quotes in the very beginning and in the very end are for clarity and are not a part of f1.

It can be seen that the characters in fi are letters, question marks, (possibly) quotation marks and spaces.

Nephren will ask the little leprechauns q times. Each time she will let them find the k-th character of fn. The characters are indexed starting from 1. If fn consists of less than k characters, output '.' (without quotes).

Can you answer her queries?

Input

The first line contains one integer q (1 ≤ q ≤ 10) — the number of Nephren's questions.

Each of the next q lines describes Nephren's question and contains two integers n and k (0 ≤ n ≤ 105, 1 ≤ k ≤ 1018).

Output

One line containing q characters. The i-th character in it should be the answer for the i-th query.

Examples

input

3
1 1
1 2
1 111111111111

output

Wh.

input

5
0 69
1 194
1 139
0 47
1 66

output

abdef

input

10
4 1825
3 75
3 530
4 1829
4 1651
3 187
4 584
4 255
4 774
2 474

output

Areyoubusy

Note

For the first two examples, refer to f0 and f1 given in the legend.

思路： 字符串长度的递推式：f[n]=2*f[n-1]+l1+l2+l3，然后分成5段来找所求字母，l1-(s[n-1])-l2-(s[n-1])-l3. 

代码：

#include<bits/stdc++.h>
#define db double
#define ll long long
#define vec vector<ll>
#define Mt  vector<vec>
#define ci(x) scanf("%d",&x)
#define cd(x) scanf("%lf",&x)
#define cl(x) scanf("%lld",&x)
#define pi(x) printf("%d
",x)
#define pd(x) printf("%f
",x)
#define pl(x) printf("%lld
",x)
//#define rep(i, x, y) for(int i=x;i<=y;i++)
const int N   = 1e6 + 5;
const int mod = 1e9 + 7;
const int MOD = mod - 1;
const db  eps = 1e-18;
const db  PI  = acos(-1.0);
using namespace std;
string s0="What are you doing at the end of the world? Are you busy? Will you save us?",
s1="What are you doing while sending "",s2=""? Are you busy? Will you send "",s3=""?";

ll f[N]={75,218,504,1076,2220};
int l1,l2,l3;
void init()
{
    for(int i=5;;i++){
        ll x=f[i-1]*2+68;
        if(x>1e18) break;
        f[i]=f[i-1]*2+68;
    }
}
char dfs(ll n,ll k)
{
    if(!n) return k<=75?s0[k-1]:'.';
    if(k<=l1) return s1[k-1];//第一段
    k-=l1;
    if(k<=f[n-1]||!f[n-1]) return dfs(n-1,k);//在范围内或者当前字符串长度远超目标位置
    k-=f[n-1];
    if(k<=l2) return s2[k-1];//s2的范围内
    k-=l2;
    if(k<=f[n-1]||!f[n-1]) return dfs(n-1,k);
    k-=f[n-1];
    return k<=l3?s3[k-1]:'.';//最后一段

}
int main(){
    init();
    ll n,k;
    int t;
    ci(t);
    l1=(int)s1.size(),l2=(int)s2.size(),l3=(int)s3.size();
    while(t--)
    {
        cl(n),cl(k);
        putchar(dfs(n,k));
    }
    return 0;
}