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#include<bits/stdc++.h>
#define ALL(X)        X.begin(),X.end()
#define FOR(I,A,B)    for(int (I) = (A); (I) <= (B); (I)++)
#define FORW(I,A,B)   for(int (I) = (A); (I) < (B);  (I)++)
#define FORD(I,A,B)   for(int (I) = (A); (I) >= (B); (I)--)
#define CLEAR(X)      memset(X,0,sizeof(X))
#define SIZE(X)       int(X.size())
#define CONTAINS(A,X) (A.find(X) != A.end())
#define PB            push_back
#define MP            make_pair
#define X             first
#define Y             second
using namespace std;
template<typename T, typename U> ostream& operator << (ostream& os, const pair<T, U> &_p) { return os << "(" << _p.X << "," << _p.Y << ")"; }
template<typename T> ostream& operator << (ostream &os, const vector<T>& _V) { bool f = true; os << "["; for(auto v: _V) { os << (f ? "" : ",") << v; f = false; } return os << "]"; }
template<typename T> ostream& operator << (ostream &os, const set<T>& _S) { bool f = true; os << "("; for(auto s: _S) { os << (f ? "" : ",") << s; f = false; } return os << ")"; }
template<typename T, typename U> ostream& operator << (ostream &os, const map<T, U>& _M) { return os << set<pair<T, U>>(ALL(_M)); }
typedef signed long long slong;
typedef long double ldouble;
typedef pair<int,int> pii;
const slong INF = 1000000100;
const ldouble EPS = 1e-9;

template<class TH> void _dbg(const char *sdbg, TH h){cerr<<sdbg<<"="<<h<<"\n";}
template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {
  while(*sdbg!=',')cerr<<*sdbg++;cerr<<"="<<h<<","; _dbg(sdbg+1, a...);
}

#ifdef LOCALs
#define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
#else
#define debug(...) (__VA_ARGS__)
#define cerr if(0)cout
#endif

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<typename X, typename Y>
using ordered_map = tree<X, Y, less<X>, rb_tree_tag, tree_order_statistics_node_update>;
//public methods: order_of_key, find_by_order

int MAXN = 300000;
int MAXK = 10; /// ??????????????
const slong inf = INF * INF;

typedef vector<slong> vl;
vector<vl> dp;

void prep() {
    dp.clear();
    dp.resize(MAXN+2,vl(MAXK+2,0));
    dp[0][0] = 1;
    FORW(n,0,MAXN) {
        FOR(i,0,MAXK) {
            FOR(j,0,min(MAXK-i,n)) {
                dp[n+1][i+j] += dp[n][i];
                if(dp[n+1][i+j] > inf) {
                    dp[n+1][i+j] = inf;
                }
            }
        }
    }
}

template<typename T>
struct tree {
    T *A;
    int M;
    function<T(T, T)> f;
    T one;

    tree(int _n, T _one, function<T(T, T)> _f) {
        M = 1 << (32 - __builtin_clz(_n));
        A = new T[2*M];
        f = _f;
        FORW(i,0,2*M) A[i] = _one;
    }

    T query(int a, int b) {
        a += M;
        b += M;
        T ret = A[a];
        if(a != b) ret = f(ret, A[b]);
        while(a/2 != b/2) {
            if(a%2 == 0) ret = f(ret, A[a+1]);
            if(b%2 == 1) ret = f(ret, A[b-1]);
            a /= 2;
            b /= 2;
        }
        return ret;
    }

    void insert(int a, T v) {
        a += M;
        A[a] = v;
        a /= 2;
        while(a != 0) {
            A[a] = f(A[2*a], A[2*a+1]);
            a /= 2;
        }
    }
};

int pl(int a, int b) { return a + b;}

vector<int> solve2(int n, slong k, int cnt) {
    /// k-ta leksykograficznie permutacja o cnt porzadkach
    vector<int> ret;
    debug(n,k,cnt);
    if(n == 1) {
        if(k == 1 && cnt == 0) {
            ret.PB(1);
        }
        assert(k == 1 && cnt == 0);
        return ret;
    }

    ordered_map<int,int> S;
    FOR(i,0,n) S[i] = i;

    while(n > 0) {
        debug(n, k, cnt);
        FOR(i,max(1,n-cnt),n) {
            int cnt2 = cnt - (n - i);
            assert(cnt2 >= 0);
            if(dp[n-1][cnt2] >= k) {
                int val = S.find_by_order(i)->X;
                ret.PB(val);
                S.erase(val);
                cnt = cnt2;
                debug(n, i);
                break;
            }
            else {
                k -= dp[n-1][cnt2];
            }
        }
        n--;
    }
    debug(ret);
    return ret;
}

slong sqn(slong n) {
    return n * (n - 1) / 2;
}

void solve()
{
    slong n, k;
    cin >> n >> k;
    MAXN = n + 2;
    MAXK = min(slong(sqrt(200000000LL / MAXN)), sqn(n) + 1);
    prep();
    debug(MAXN,MAXK,dp[MAXN][MAXK]);
    debug(dp[3][1]);
    //slong ko = k;

    slong tg_inv = sqn(n);
    int fail = 0;
    if(tg_inv % 2) fail = 1;
    tg_inv /= 2;

    if(!fail) {
        slong i = n;
        slong m = n - i + 1;
        while(sqn(m) < tg_inv) {
            i--;
            m = n - i + 1;
        }
        slong rem = sqn(m) - tg_inv; /// ile inwersji trzeba zabrac
        slong ai = n - rem;
        cerr << "start: "; debug(i, ai);
        while(i > 0) {
            m = n - i + 1; /// ostatnie m zaczyna sie od ai
            rem = sqn(m) - (n - ai) - tg_inv;
            assert(rem >= 0);
            debug(k, i, ai, rem, m, dp[m-1][rem]);
            if(dp[m-1][rem] < k) {
                k -= dp[m-1][rem];
                ai++;
                if(ai > n) {
                    i--;
                    ai = i + 1;
                }
            }
            else {
                debug(i, ai);
                vl ans(i);
                FORW(j,0,i) ans[j] = j + 1;
                ans[i-1] = ai;
                auto suf = solve2(m-1, k, rem);
                for(int x : suf) ans.PB(i - 1 + x + (x + i - 1 >= ai)); /// POPRAWIC >= !!!!!!!!!!!!!!
                cout << "TAK\n";
                for(slong x : ans) cout << x << " ";
                cout << "\n";
                return;
            }
        }
    }
    cout << "NIE\n";
}

int main()
{
    ios_base::sync_with_stdio(0);
    solve();
}