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#include <iostream>
#include <vector>
#include <set>
#include <limits>
using namespace std;

void solve_k2(vector<int>& V) {
    int n = V.size();

    vector<int> minL(n);
    vector<int> maxR(n);
    minL[0] = V[0];
    maxR[n-1] = V[n-1];

    for (int i = 1; i < n; i++)
        minL[i] = min(minL[i-1], V[i]);
    for (int i = n - 2; i >= 0; i--)
        maxR[i] = max(maxR[i+1], V[i]);

    for (int i = 0; i < n - 1; i++) {
        if (minL[i] >= maxR[i+1]) {
            cout << "TAK\n";
            cout << i + 1 << "\n";
            return;
        }
    }

    cout << "NIE\n";
}

void solve_k3(vector<int>& V) {
    int n = V.size();

    int mn = numeric_limits<int>::max(), mx = numeric_limits<int>::min();
    for (int i = 0; i < n; i++) {
        mn = min(mn, V[i]);
        mx = max(mx, V[i]);
    }

    for (int i = 1; i < n; i++)
        if (V[i] == mn) {
            cout << "TAK\n";
            if (i == n - 1)
                cout << i - 1 << " " << i << "\n";
            else
                cout << i << " " << i + 1 << "\n";
            return;
        }
    
    for (int i = 0; i < n - 1; i++)
        if (V[i] == mx) {
            cout << "TAK\n";
            cout << i << " " << i + 1 << "\n";
            return;
        }

    cout << "NIE\n";
}

void solve_kbig(vector<int>& V, int k) {
    int n = V.size();

    // find the pair
    int pair_pos;
    for (pair_pos = 0; pair_pos + 1 < n; pair_pos++)
        if (V[pair_pos] >= V[pair_pos + 1])
            break;
    if (pair_pos + 1 == n) {
        cout << "NIE\n";
        return;
    }

    // insert initial ranges
    set<int>Result;
    if (pair_pos != 0)
        Result.insert(pair_pos - 1);
    Result.insert(pair_pos);
    Result.insert(pair_pos + 1);

    // insert the rest arbitiary
    for (int i = n - 1; (int)Result.size() < k; i--)
        Result.insert(i);
    Result.erase(prev(Result.end()));

    // output
    cout << "TAK\n";
    for (int x : Result)
        cout << x + 1 << " ";
    cout << "\n";
}

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

    int n, k;
    cin >> n >> k;
    vector<int>A(n);
    for (int &x : A)
        cin >> x;

    if (k == 2)
        solve_k2(A);
    else if (k == 3)
        solve_k3(A);
    else
        solve_kbig(A, k);

    return 0;
}