//~~~~~~~~~~~~~MJ®™~~~~~~~~~~~~~
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#define rep1(n) for(ll i=0; i<(ll)(n); ++i)
#define rep2(i,n) for(ll i=0; i<(ll)(n); ++i)
#define rep3(i,a,b) for(ll i=(ll)(a); i<(ll)(b); ++i)
#define rep4(i,a,b,c) for(ll i=(ll)(a); i<(ll)(b); i+=(c))
#define cut4(a,b,c,d,e,...) e
#define rep(...) cut4(__VA_ARGS__,rep4,rep3,rep2,rep1)(__VA_ARGS__)
#define per1(n) for(ll i=((ll)n)-1; i>=0; --i)
#define per2(i,n) for(ll i=((ll)n)-1; i>=0; --i)
#define per3(i,a,b) for(ll i=((ll)a)-1; i>=(ll)(b); --i)
#define per4(i,a,b,c) for(ll i=((ll)a)-1; i>=(ll)(b); i-=(c))
#define per(...) cut4(__VA_ARGS__,per4,per3,per2,per1)(__VA_ARGS__)
#define ll long long
#define ln cout<<endl
#define int long long
#define vv vector<vi>
#define vp vector<pi>
#define append push_back
#define all(x) (x).begin(),(x).end()
#define allr(x) (x).rbegin(),(x).rend()
#define vi vector<int>
#define ret(x) {cout<<x;return;}
#define ui map<int,int>
#define pi pair<int,int>
#define ff first
#define ss second
using namespace std;
const int INF = 1e18, MOD = 1e9+7, N = 9e6+7;
int binary_to_int(string s) {int n = 0; for (int i = 0; i < s.size(); i++) {n = n * 2 + s[i] - '0';} return n;}
// ul graph[N];
bool vis[N];
bool having[N];
vi path;
int st,en;
int n;
// void print()
// {
// for (int i:path)
// {
// cout << i << " ";
// }
// ln;
// }
// void dfs(int node ,int p)
// {
// vis[node]=1;
// // path.append(node);
// for (int j= n-1; j>=0; j--)
// {
// int i = 1<<j ^ node;
// if (!having[i] and !vis[i])
// {
// dfs(i,node);
// }
// }
// // path.pop_back();
// }
void bfs(int node, int p)
{
queue<int> q;
q.push(node);
vis[node]=1;
while (!q.empty())
{
int node = q.front();
q.pop();
for (int j= n-1; j>=0; j--)
{
int i = 1<<j ^ node;
if (!having[i] and !vis[i])
{
vis[i]=1;
q.push(i);
}
}
}
}
void solve()
{
int k, m, ans = 0;
string e, f;
cin >> n >> k >> e >> f;
st = binary_to_int(e);
en = binary_to_int(f);
vi a(k);
rep(k)
{
cin >> e;
a[i]=binary_to_int(e);
having [a[i]]=1;
}
bfs(st,-1);
if (vis[en])
{
cout <<"TAK";
}
else
{
cout <<"NIE";
}
// cout << ans;
// cout << a.size();
// for (auto i: a){cout << i << " ";}
}
signed main(){
ios_base::sync_with_stdio(0);
cin.tie(NULL); cout.tie(NULL);
int ans=1;
//cout<<setprecision(1000);
// cin>>ans;
rep(ans){
// cout << "Case #" << i+1 << ": ";
solve();ln;}}
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