#include <iostream>
#include <algorithm>

using namespace std;

typedef int number;
#define MAXN 250000
#define EXIT_GRACEFULLY delete[] temp;\
                        return 0;

bool odd(number i);

int log(int n);
int licz[MAXN];

//Rytter algorithm
int inversionNumber(number perm[], int n) {
    int result = 0;

    for (int i = 0; i < log(n); ++i) {
        for (int j = 0; j < n; ++j) {
            if(odd(perm[j])) {
                licz[perm[j]] += 1;
            } else {
                result += licz[perm[j] + 1];
            }
        }
        for (int j = 0; j < n; ++j) {
            licz[j] = 0;
            perm[j] /= 2;
        }
    }

    return result;
}

int pairNumber(int n) {
    return n * (n-1) / 2;
}

int log(int n) {
    char exponent = 0;
    while(n != 0) {
        n >>= 1;
        exponent++;
    }
    return exponent;
}

bool odd(number i) {
    return i & 1;
}

void print(number pInt[], int n) {
    for (int i = 0; i < n; ++i) {
        cout << pInt[i] << " ";
    }
    cout << endl;
}

int main() {
    int n;
    long long k = 1;
    long long kDesired;
    number perm[MAXN];
    cin >> n >> kDesired;
    const int pn = pairNumber(n) / 2;
    number *temp = new number[n];
    iota(perm, perm+n, 1);
    bool found = true;
    do {
        copy(perm, perm+n, temp);
        if(pn == inversionNumber(temp, n)) {
            if(k == kDesired) {
                cout << "TAK" << endl;
                print(perm, n);
                break;
            }
            k++;
        }

    } while (found = next_permutation(perm, perm+n));
    if(!found) {
        cout << "NIE" << endl;
    }

    EXIT_GRACEFULLY
}