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# include <cstdio>
# include <cstring>
# include <algorithm>
# include <iostream>
using namespace std ;
const int N = 1e6 + 10 ;
int n ; //模式串数量
char s [ N ] ; //模式串
// Trie树
int tr [ N ] [ 2 ] , idx ;
int cnt [ N ] ;
void insert ( char * s ) {
int p = 0 ;
for ( int i = 0 ; s [ i ] ; i + + ) {
int t = s [ i ] - ' 0 ' ;
if ( ! tr [ p ] [ t ] ) tr [ p ] [ t ] = + + idx ;
p = tr [ p ] [ t ] ;
}
cnt [ p ] + + ; //记录以p点为结尾的模式串数量+1, 也就是说它是危险节点
}
// AC自动机
int q [ N ] , ne [ N ] ;
void bfs ( ) {
int hh = 0 , tt = - 1 ;
for ( int i = 0 ; i < 2 ; i + + ) //将第一层存在的节点入队列
if ( tr [ 0 ] [ i ] ) q [ + + tt ] = tr [ 0 ] [ i ] ;
while ( hh < = tt ) {
int p = q [ hh + + ] ;
for ( int i = 0 ; i < 2 ; i + + ) {
int t = tr [ p ] [ i ] ;
if ( ! t )
tr [ p ] [ i ] = tr [ ne [ p ] ] [ i ] ;
else {
ne [ t ] = tr [ ne [ p ] ] [ i ] ;
q [ + + tt ] = t ;
// tr[t][i]这个节点,它的失配指针指向的节点,也是危险节点的话,那么,当前节点也是危险节点
if ( cnt [ ne [ t ] ] ) cnt [ t ] + + ;
}
}
}
}
// dfs在trie树上判环的代码模板
int st [ N ] ;
bool dfs ( int u ) { //在AC自动机上判环
if ( st [ u ] = = 1 ) return true ; //如果在判环过程中发现重复访问的点,说明存在环
if ( st [ u ] = = - 1 ) return false ; //如果以前检查过这个点u,结果一条路跑过黑, 都没有检查出来过有环, 那么下次如果再让我检查点u,就直接返回结果就可以了
st [ u ] = 1 ; //当前路径上走过了点u
for ( int i = 0 ; i < = 1 ; i + + ) { //二进制, 必然每个节点有两条分叉边, 0和1
if ( ! cnt [ tr [ u ] [ i ] ] ) { //如果tr[u][i]这个点是危险点的话,就避让开;如果不是危险点的话,就继续探索它
if ( dfs ( tr [ u ] [ i ] ) ) return true ; // 如果后续发现存在环,那么我也可以上报信息:我的后代子孙中发现环,也就是我这一脉中存在环;
//如果tr[u][i]及后续不存在环,那还需要继续检查,不能直接返回 dfs(tr[u][i]),这一块要注意写法,也算是一种代码模板,需要背诵下来
}
}
st [ u ] = - 1 ; //一条道跑到黑,都没有找出后续节点中存在环,那么标识上,防止以后重复查询本节点
return false ; //如果找到了环, 中途就会返回true,都跑到这里了,表示没有找到环
}
int main ( ) {
//加快读入
ios : : sync_with_stdio ( false ) , cin . tie ( 0 ) ;
cin > > n ;
for ( int i = 1 ; i < = n ; i + + ) {
cin > > s ;
insert ( s ) ;
}
//构建AC自动机
bfs ( ) ;
//从root出发, 开始判断是不是存在环
dfs ( 0 ) ? puts ( " TAK " ) : puts ( " NIE " ) ;
return 0 ;
}