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// Krzysztof Małysa
#include <bits/stdc++.h>
using namespace std;

#define FOR(i,a,n) for (auto i = (a), i ## __ = (n); i <= i ## __; ++i)
#define FORD(i,a,n) for (auto i = (a), i ## __ = (n); i >= i ## __; --i)
#define REP(i,n) FOR(i, 0, n - 1)
#define ALL(x) x.begin(), x.end()
#define SZ(x) (int(x.size()))
#define EB emplace_back
#define ST first
#define ND second
#define tpv typedef vector<

typedef long long LL;
typedef pair<int, int> PII;
tpv int> VI;
tpv VI> VVI;
tpv PII> VPII;
tpv LL> VLL;

constexpr char nl = '\n';
#define endl nl

#define ris return *this
#define tem template<class T

tem, class B> inline void mini(T&& a, B&& b) { if (b < a) a = b; }
tem, class B> inline void maxi(T&& a, B&& b) { if (b > a) a = b; }
int ceil2(int x) { return x < 2 ? 1 : 1 << (sizeof(x) * 8 - __builtin_clz(x - 1)); }

tem> struct Dump { T a, b; };
tem> auto dump(T&& x) -> Dump<decltype(x.begin())> { return {ALL(x)}; }
struct Debug {
	~Debug() { cerr << endl; }
	tem> auto operator<<(T x) -> decltype(cerr << x, *this) { cerr << boolalpha << x; return *this; }
	tem> auto operator<<(T x) -> decltype(x.begin(), *this) {
		auto a = x.begin(), b = x.end();
		*this << "{";
		for (; a != b;)
			*this << *a++, *this << (a == b ? "" : " ");
		return *this << "}";
	}
	tem, class B> Debug& operator<<(pair<T, B> p) { ris << "(" << p.ST << ", " << p.ND << ")"; }
	tem> Debug& operator<<(Dump<T> d) {
		*this << "{\n";
		for (T a = d.a, c = a; a != d.b; ++a)
			*this << "  " << distance(c, a) << ": " << *a << '\n';
		ris << "}";
	}
};
struct Foo {tem>Foo& operator<<(T) {ris;}};

#ifdef DEBUG
# define deb Debug()
#else
# define deb Foo()
#endif
#define imie(x...) #x ": " << (x) << " "
#define LOG(x...) deb << imie(x)
#define DOG(x...) deb << #x ": " << dump(x) << " "

class ITree {
	int size;
	VI t; // t[i] - how many zeros are in the subtree of i

public:
	explicit ITree(int n) : size(ceil2(n)), t(size << 1, 1) {
		FORD (i, size - 1, 1)
			t[i] = t[i << 1] + t[i << 1 | 1];
	}

	void set1(int x) {
		x += size;
		if (t[x]) {
			t[x] = 0;
			while (x >>= 1)
				--t[x];
		}
	}

	// Finds k-th 0, assuming that first is 0-th
	int findkth0(int k) { return findkth0(k, 1, size); }

	int findkth0(int k, int i, int sz) {
		LOG(VI{k, i, sz});
		if (sz == 1)
			return i - size;

		int sz2 = sz >> 1;
		i <<= 1;
		if (t[i] <= k)
			return findkth0(k - t[i], i + 1, sz2);
		else
			return findkth0(k, i, sz2);
	}

};

const LL INF = 1000000000000000001LL;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int n;
	LL k;
	cin >> n >> k;
	// Calculate dp - dp[i][j] = # of permutations of length i having j inversions
	vector<vector<LL>> dpp(2);
	auto dp = [&](int i, LL j, LL sz) {
		if (j > sz)
			return 0LL;

		j = min(j, sz - j);
		if (j >= SZ(dpp[i]))
			return INF;
		else
			return dpp[i][j];
	};

	dpp[0] = {};
	dpp[1] = {1};
	FOR (i, 2, n) {
		dpp.EB(VLL{1});
		FOR (j, 1LL, i * LL(i - 1) >> 2) {
			LL dppij = dpp[i][j - 1];
			LL sz = LL(i - 2) * (i - 1) >> 1;
			dppij += dp(i - 1, j, sz);
			if (j >= i)
				dppij -= dp(i - 1, j - i, sz);

			if (dppij >= INF)
				break;
			dpp[i].EB(dppij);
		}
	}

	DOG(dpp);

	// Find the result
	LL j = LL(n - 1) * n >> 2;
	if (LL(n - 1) * n & 3 || dp(n, j, LL(n - 1) * n >> 1) < k)
		return puts("NIE"), 0;

	cout << "TAK\n";
	--k;
	ITree tree(n);
	LL sum = 0;
	FORD (i, n, 2) {
		LL p = min(LL(i - 2) * (i - 1) >> 1, j);
		LL sz = LL(i - 2) * (i - 1) >> 1;
		for (;;) {
			LL x = dp(i - 1, p, sz);
			// cerr << imie(x) << nl;
			if (x > k)
				break;

			k -= x;
			--p;
			++sum;
		}

		// LOG(VLL{i, p, j, k});
		LOG(j - p);
		int x = tree.findkth0(j - p);
		LOG(x);
		tree.set1(x);
		cout << x + 1 << ' ';
		j = p;
	}

	cout << tree.findkth0(0) + 1 << nl;
	// cerr << sum << nl;
	return 0;
}