Migrated from Lutece 1269 ZhangYu Speech

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Description

as we all know, ZhangYu(Octopus vulgaris) brother has a very famous speech - "Keep some distance from me". ZhangYu brother is so rich that everyone want to contact he, and scfkcf is one of them. One day , ZhangYu brother agreed with scfkcf to contact him if scfkcf could beat him. There are $n$ digits(lets give them indices from $1$ to $n$ and name them $a_1, a_2 ... a_N$) and some queries.

for each query:

1. ZhangYu brother choose an index $x$ from $1$ to $n$.
2. For all indices $y$ ( $y$ < $x$) calculate the difference $b_y = a_x - a_y$.
3. Then ZhangYu brother calculate $B_1$ ，the sum of all by which are greater than $0$ , and scfkcf calculate $B_2$ , the sum of all by which are less than $0$.

if $B_1 > |B_2|$ , ZhangYu brother won and did not agree with scfkcf to contact him; else if $B_1$ is equals to $|B_2|$ , ZhangYu brother would ignore the result; else if $B_1$ < $|B_2|$ , ZhangYu brother lost and agreed with scfkcf to contact him.

Input

The first line contains two integers $n$, $m$ $(1 \le n,m \le 100000)$ denoting the number of digits and number of queries. The second line contains $n$ digits (without spaces) $a_1, a_2, ..., a_n$.$(0 \le a_i \le 9)$ Each of next $m$ lines contains single integer $x$ $(1 \le x \le n)$ denoting the index for current query.

Output

For each of $m$ queries print "Keep some distance from me" if ZhangYu won, else print "Next time" if ZhangYu brother ignored the result, else print "I agree" if ZhangYu brother lost in a line - answer of the query.

Samples

10 3
0324152397
1
4
7

Next time
Keep some distance from me
I agree

Note

It's better to use "scanf" instead of "cin" in your code.

Resources

第七届ACM趣味程序设计竞赛第四场（正式赛）