https://codeforces.com/contest/1151/problem/B 

题意：给定一个n∗m 的矩阵，在每一行都取一个数，问是否存在一种方法使得异或和不为0

题解：

1、随便选；

2、如果1中的结果为0；

3、暴力判断每一段有木有和已选的不一样的数；

4、如果有就就是可以（找到一行有就行了），没有就NO；

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=1000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N][N],b[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            scanf("%d",&a[i][j]);
        }
    }
    sum=0;
    for(int k=1;k<=n;k++){
        b[k]=1;
        sum^=a[k][1];
    }
    if(!sum){
        for(int i=1;i<=n;i++){
            temp=-1;
            for(int j=2;j<=m;j++){
                if(a[i][1]!=a[i][j]){
                    temp=j;
                    break;
                }
            }
            if(temp!=-1){
                b[i]=temp;
                cout<<"TAK"<<endl;
                for(int k=1;k<=n;k++)
                    printf("%d%c",b[k]," \n"[k==n]);
                return 0;
            }
        }
        cout<<"NIE"<<endl;
    }else{
        cout<<"TAK"<<endl;
        for(int i=1;i<=n;i++)
            printf("%d%c",b[i]," \n"[i==n]);
    }
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}