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#include <bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for (int i = (a); i <= (b); ++i)
#define REPD(i,a,b) for (int i = (a); i >= (b); --i)
#define FORI(i,n) REP(i,1,n)
#define FOR(i,n) REP(i,0,int(n)-1)
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define vi vector<int>
#define ll long long
#define SZ(x) int((x).size())
#define DBG(v) cerr << #v << " = " << (v) << endl;
#define FOREACH(i,t) for (typeof(t.begin()) i=t.begin(); i!=t.end(); i++)
#define fi first
#define se second

const int N = 300300;
const ll inf = 1000100100200100100LL;

vector<ll> dp[N];

const int M = 1<<18;
int T[2*M+13];

void init() {
	dp[1].pb(1);
	for (int n = 2; n < N; n++) {
		ll mx = 1LL*n*(n-1)/2+1;
		ll prv = 1LL*(n-1)*(n-2)/2+1;
		if (SZ(dp[n-1]) != prv && mx>SZ(dp[n-1])) mx=SZ(dp[n-1]);
		for (int k=0; k<mx; k++) {
			ll cur = 0;
			for (int i = 0; i < n && i <= k; i++) {
				if (SZ(dp[n-1]) > k-i) cur += dp[n-1][k-i];
				if (cur >= inf) {
					cur = inf;
					break;
				}
			}
			if (cur < inf) {
				dp[n].pb(cur);
			} else {
				break;
			}
		}
	}
	
	FOR(i,M) T[M+i]=1;
	for (int i=M-1; i>=1; i--) T[i]=T[i*2]+T[i*2+1];
}

ll calc(int n, ll m) {
	fflush(stdout);
	ll mx = 1LL*n*(n-1)/2;
	if (m>mx) return 0;
	if (m > mx-m) m=mx-m;
	if (SZ(dp[n]) <= m) return inf;
	return dp[n][m];
}

int ith(int x) {
	int p = 1, sl = 0;
	while (p < M) {
		if (sl + T[p*2] <= x) {
			p = p*2+1;
			sl += T[p-1];
		} else p = p*2;
	}
	int ret = p-M;
	T[p]--;
	while (p>1) {
		p /= 2;
		T[p]=T[p*2]+T[p*2+1];
	}
	return ret;
}

int res[N];
int main() {
	init();
	int n;
	ll k;
	scanf("%d%lld", &n, &k);
	ll m = 1LL*n*(n-1)/2;
	if (m%2==1) {
		printf("NIE\n");
		return 0;
	}
	m /= 2;
	if (SZ(dp[n]) > m && dp[n][m] < k) {
		printf("NIE\n");
		return 0;
	}
	FOR(i,n) {
		ll lo = m-1LL*(n-i-1)*(n-i-2)/2;
		if (lo<0) lo=0;
		if (lo>n) lo=n;
		for (int c=lo; c<n; c++) {
			ll cur = calc(n-i-1, m-c);
			if (k <= cur) {
				res[i] = ith(c);
				m -= c;
				break;
			}
			k -= cur;
		}
	}
	printf("TAK\n");
	FOR(i,n) printf("%d ", res[i]+1);
	printf("\n");
	return 0;
}