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#include <bits/stdc++.h>

#define FOR(i,b,e) for(int i=(b); i <= (e); ++i)
#define FORD(i,b,e) for(int i=(b); i >= (e); --i)
#define SIZE(c) (int) (c).size()
#define FORE(i,c) FOR(i,0,SIZE(c)-1)
#define FORDE(i,c) FORD(i,SIZE(c)-1,0)

#define pb push_back
#define mp make_pair
#define st first
#define nd second


using namespace std;

typedef long long ll;
typedef pair <int,int> pii;
typedef pair <ll,ll> pll;

typedef vector <int> VI;
typedef vector <bool> VB;
typedef vector <pii> VP;
typedef vector <ll> VL;
typedef vector <pll> VPL;

/*************************************************************************/

const int N = 250002;
const ll LINF = 1000000000000000001LL;

ll maxInv[N];
VL dp[N];

void prepare(int n) {
    FOR(i,0,n) {
        maxInv[i] = (ll) i * (i - 1) / 2;
        
        dp[i] = {1};
        for (ll j = 1; j <= maxInv[i]; j++) {
            dp[i].pb(0);
            
            ll firstFrom = max(0LL, j + 1 - SIZE(dp[i-1]));
            ll firstTo = min(j, (ll) i-1);
            
            for (ll first = firstFrom; first <= firstTo; first++) {
                dp[i].back() = min(dp[i].back() + dp[i-1][j-first], LINF);
            }
            
            if (dp[i].back() == LINF) {
                break;
            }
        }
    }
}

/*************************************************************************/

struct ITree {
    VI T;
    int p = 1;

    ITree(int n) {
        while (p < n) p *= 2;
        T.resize(2*p-1,0);
        
        FOR(i,0,n-1) {
            T[i + p - 1] = 1;
        }
        
        FORD(i,p-2,0) {
            T[i] = T[2 * i + 1] + T[2 * i + 2];
        }
    }

    int par(int x) { return (x - 1) / 2; }

    void del(int l) {
        l += p-1;
        T[l]--;

        while (l > 0) {
            l = par(l);
            T[l]--;
        }
    }

    int kth(int v, int k) {
        if (v >= p-1) {
            return v - p + 1;
        }
        
        int inLeft = T[2 * v + 1];
        
        if (k < inLeft) {
            return kth(2 * v + 1, k);
        } else {
            return kth(2 * v + 2, k - inLeft);
        }
    }
    
    int kth(int k) { return kth(0, k); }
};

/*************************************************************************/

void printYes(VI &ans) {
    cout << "TAK\n";

    for (int x : ans) {
        cout << 1 + x << ' ';
    }
}

void printNo() { cout << "NIE"; }

bool solve(int n, ll k, ll inv, VI &ans) {
    ITree tree(n);

    FOR(i,0,n-1) {
        int len = n - i;
        
        int found = -1;
        ll firstFrom = max(0LL, inv - maxInv[len-1]);
        ll firstTo = min((ll) len-1, inv);
        
        for (ll first = firstFrom; first <= firstTo; first++) {
            ll left = inv - first;
            left = min(left, maxInv[len-1] - left);
            
            ll here;
            if (left < SIZE(dp[len-1])) {
                here = dp[len-1][left];
            } else {
                here = LINF;
            }
            
            if (here >= k) {
                found = first;
                break;
            }
            
            k -= here;
        }
        
        if (found == -1) {
            return false;
        }
        
        inv -= found;
        
        int v = tree.kth(found);
        tree.del(v);
        
        ans.pb(v);
    }
    
    return true;
}

/*************************************************************************/

int main() {
    ios_base::sync_with_stdio(0);
    
    int n;
    ll k;
    
    cin >> n >> k;
    prepare(n);
    
    if (maxInv[n] % 2) {
        printNo();
        return 0;
    }
    
    VI ans;
    
    if (solve(n, k, maxInv[n] / 2, ans)) {
        printYes(ans);
    } else {
        printNo();
    }

    return 0;
}

/*************************************************************************/