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#include<bits/stdc++.h>
#define PII pair<int,int>
#define f first
#define s second
#define VI vector<int>
#define LL long long
#define MP make_pair
#define LD long double
#define PB push_back
#define ALL(V) V.begin(),V.end()
#define abs(x) max((x),-(x))
#define PDD pair<LD,LD> 
#define VPII vector< PII > 
#define siz(V) ((int)V.size())
#define FOR(x, b, e)  for(int x=b;x<=(e);x++)
#define FORD(x, b, e) for(int x=b;x>=(e);x--)
#define REP(x, n)     for(int x=0;x<(int)(n);x++)
#define mini(a,b) a=min(a,b)
#define maxi(a,b) a=max(a,b)
using namespace std;
const int MXN = 250100;

vector<LL> dp[MXN];//dp[n][k] = ile permutacji liczb 1...n mających k inwersji
vector<bool> memo[MXN];
LL inf = 1.1e18;

LL go(LL n, LL k)
    {
    if(k < 0)return 0;
    if(n == 0 && k == 0)return 1;
    if(n == 0)return 0;
    LL po2 = (LL)n * (n-1) / 2;
    if(k > po2)return 0;
    k = min(k, po2-k);

    if(k >= (int)memo[n].size())return inf;
    if(memo[n][k])return dp[n][k];
    memo[n][k] = 1;
    
    FOR(i, 1, n)
        {
        int new_inv = i-1;
        
        if(k - new_inv < 0)break; // po break mało operacji (353551)
        dp[n][k] += go(n-1, k - new_inv);

        if(dp[n][k] >= inf)
            {
            dp[n][k] = inf;
            return inf;
            }
        }
    return dp[n][k];
    }
    

VI get(LL n, LL inv, LL kth)
    {
    VI V;
    FOR(_, 1, n)
        {
        LL maxinv = (n-_) * (n-_ - 1) / 2;
        
        FOR(i, max(1LL, inv-maxinv+1), n)  // działa szybko bo max
            {
            LL ways = go(n-_, inv-(i-1));
            if(ways < kth)
                {
                kth -= ways;
                }
            else
                {
                V.PB(i);
                inv -= (i-1);
                break;
                }
            }
        }
    return V;
    }
    
#include<ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
VI repair(VI V)
    {
    ordered_set<int> S;
    FOR(i, 1, (int)V.size())
        S.insert(i);
    
    VI V2;
    REP(i, V.size())
        {
        int x = *S.find_by_order(V[i]-1);
        
        S.erase(x);
        V2.PB(x);
        }
    return V2;
    }
    
int main()
    {
    LL n, k;
    cin>>n>>k;
    
    
    REP(i, 1000)
        {
        dp[i].resize(100);
        memo[i].resize(100);
        }
    FOR(i, 1000, n)
        {
        dp[i].resize(10);
        memo[i].resize(10);
        }
    
    LL pol = n * (n-1) / 4;
    
    if((n * (n-1)/2) % 2 != 0 || go(n, pol) < k)
        {
        puts("NIE");
        return 0;
        }
    
    VI V = get(n, pol, k);
    V = repair(V);
    
    puts("TAK");
    for(auto i : V)
        printf("%d ", i); 
    puts("");
    }