CF1151B.Dima and a Bad XOR

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题目描述

Student Dima from Kremland has a matrix aa of size n×mn \times m filled with non-negative integers.

He wants to select exactly one integer from each row of the matrix so that the bitwise exclusive OR of the selected integers is strictly greater than zero. Help him!

Formally, he wants to choose an integers sequence c1,c2,,cnc_1, c_2, \ldots, c_n ( 1cjm1 \leq c_j \leq m ) so that the inequality a1,c1a2,c2an,cn>0a_{1, c_1} \oplus a_{2, c_2} \oplus \ldots \oplus a_{n, c_n} > 0 holds, where ai,ja_{i, j} is the matrix element from the ii -th row and the jj -th column.

Here xyx \oplus y denotes the bitwise XOR operation of integers xx and yy .

输入格式

The first line contains two integers nn and mm ( 1n,m5001 \leq n, m \leq 500 ) — the number of rows and the number of columns in the matrix aa .

Each of the next nn lines contains mm integers: the jj -th integer in the ii -th line is the jj -th element of the ii -th row of the matrix aa , i.e. ai,ja_{i, j} ( 0ai,j10230 \leq a_{i, j} \leq 1023 ).

输出格式

If there is no way to choose one integer from each row so that their bitwise exclusive OR is strictly greater than zero, print "NIE".

Otherwise print "TAK" in the first line, in the next line print nn integers c1,c2,cnc_1, c_2, \ldots c_n ( 1cjm1 \leq c_j \leq m ), so that the inequality a1,c1a2,c2an,cn>0a_{1, c_1} \oplus a_{2, c_2} \oplus \ldots \oplus a_{n, c_n} > 0 holds.

If there is more than one possible answer, you may output any.

输入输出样例

  • 输入#1

    3 2
    0 0
    0 0
    0 0
    

    输出#1

    NIE
    
  • 输入#2

    2 3
    7 7 7
    7 7 10
    

    输出#2

    TAK
    1 3 
    

说明/提示

In the first example, all the numbers in the matrix are 00 , so it is impossible to select one number in each row of the table so that their bitwise exclusive OR is strictly greater than zero.

In the second example, the selected numbers are 77 (the first number in the first line) and 1010 (the third number in the second line), 710=137 \oplus 10 = 13 , 1313 is more than 00 , so the answer is found.

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