import std.stdio, std.conv, std.string, std.array, std.range, std.algorithm, std.container;
import std.math, std.random, std.bigint, std.datetime, std.format;
void main(string[] args){ if(args.length > 1) if(args[1] == "-debug") DEBUG = 1; solve(); }
void log()(){ writeln(""); } void log(T, A ...)(T t, lazy A a){ if(DEBUG) write(t, " "), log(a); } bool DEBUG = 0; 
string read(){ static string[] ss; while(!ss.length) ss = readln.chomp.split; string res = ss[0]; ss.popFront; return res; }
// ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- ----- //
void solve(){
	int n = read.to!int;
	int k = read.to!int;
	
	char[] as = readln.chomp.to!(char[]);
	
	int l0, r0, l1, r1;
	bool f0, f1;
	foreach(int i, a; as){
		if(a == '1'){
			if(! f0) l0 = i, f0 = 1;
			r0 = i;
		}
		else{
			if(! f1) l1 = i, f1 = 1;
			r1 = i;
		}
	}
	
	if(r0 - l0 + 1 <= k || r1 - l1 + 1 <= k){
		// can take in one move
		"tokitsukaze".writeln;
		return;
	}
	if(r0 - l0 + 1 >= k + 2 || r1 - l1 + 1 >= k + 2){
		// flips {l + 1, ..., r - 1} or so
		"once again".writeln;
		return;
	}
	
	if(l0 <= k - 1 && r0 >= n - k || l1 <= k - 1 && r1 >= n - k){
		// First cannot avoid taking one stone
		"quailty".writeln;
		return;
	}
	
	"once again".writeln;
	
	
}
/*
Second can always mimic the First. So,

- First wins at his first move.
- Whatever move First plays, Second wins at the next move.

*/
