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#include <bits/stdc++.h>
#include <chrono>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace chrono;
using namespace __gnu_pbds;
#pragma GCC optimize("Ofast,unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma,tune=native")
//#pragma GCC target("sse,sse2,sse3,mmx,abm,tune=native") // for szkopul and sio only
typedef long long lld;
typedef double lf;
typedef long double llf;
typedef pair<int,int> pii;
typedef pair<lld,lld> pll;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
#define For(i,s,a) for(lld i = (lld)s; i < (lld)a; ++i)
#define rpt(s, it) for(auto it = s.begin(); it != s.end(); ++it)
#define brpt(s, it) for(auto it = s.rend(); it != s.rbegin(); --it)
#define pb push_back
#define eb emplace_back
#define ff first
#define dd second
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define make_unique(x) (x).erase( unique(all(x)), (x).end())
#define popcnt(x) __builtin_popcount(x)
#define sz size()
#define time_since duration_cast< milliseconds >(system_clock::now().time_since_epoch())
 
template<typename Ta, typename Tb>
ostream & operator <<(ostream & os, pair<Ta, Tb> x){
    return os << x.ff << " " << x.dd;
}

const int N = 1e6 + 1;
int c[N], mn[N], mx[N];

void solve_2(int n) {
    For (i, 0, n - 1)
        if (mn[i] >= mx[i + 1]) {
            printf("TAK\n%d", i + 1);
            return;
        }
    puts("NIE");
}

void solve_3(int n) {
    For (i, 0, n - 1) {
        if (mn[i] >= c[i + 1]) {
            printf("TAK\n%d %d\n", i + 1, i + 2);
            return;
        }
        if (c[i + 1] >= mx[i + 2]) {
            printf("TAK\n%d %d\n", i + 1, i + 2);
            return;
        } 
    }
    puts("NIE");
}

void solve(int n, int k) {
    For (i, 0, n - 1) {
        if (c[i] >= c[i + 1]) {
            puts("TAK");
            vector<int> ans;
            ans.pb(i + 1);
            ans.pb(i + 2);
            For (j, 0, min(k - 3, (int)i))
                ans.pb(i + 1 - (j + 1));
            For (j, i + 2, k - 1)
                ans.pb(j + 1);
            sort(ans.begin(), ans.end());
            for (auto s : ans)
                printf("%d ", s);
            puts("");
            return;
        }
    }
    puts("NIE");
}

int main(void) {
    int n, k;
    scanf("%d%d", &n, &k);
    For (i, 0, n)
        scanf("%d", &c[i]);
    
    mn[0] = c[0];
    mx[n - 1] = c[n - 1];
    For (i, 1, n) {
        mn[i] = min(mn[i - 1], c[i]);
        mx[n - i - 1] = max(mx[n - i], c[n - i - 1]);
    }

    switch(k) {
        case 2:
            solve_2(n);
            break;
        case 3: 
            solve_3(n);
            break;
        default:
            solve(n, k);
    }
}