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문제

Parity is an important concept in data transmission. Because the process is not error proof, parity is used to provide a check on whether or not data has been corrupted in transmission.

If even parity is being used, each byte will have an even number of 1 characters. If odd parity is being used, each byte will have an odd number of 1 characters.

In this problem, we are looking for a single bit that has been corrupted. To help you find it, the last byte is not part of the data being transmitted, but is a parity byte. Each bit of the parity byte will be used to make the corresponding bits in the data bytes odd or even depending on the parity being used.

입력

The first line of input is a single integer, N (3 <= N <= 10), the number of bytes of data to follow. The next N lines each contain a single byte of data consisting of 8 characters separated by spaces. Each character will be either 1 or 0. 

There will be one further line of 8 characters (again 1 or 0) which will be the parity byte. In the parity byte, each bit is a parity bit for the corresponding bits in the preceding N lines, using the same parity as is used by the data bytes. The parity byte itself may not show the same parity as the data bytes.

출력

Output 3 lines of information about the data in the input. 

  • Line 1: Either the word Even or the word Odd to describe the parity being used by the bytes which are not broken.
  • Line 2: Byte is broken
  • Line 3: Bit is broken

is the number of the appropriate byte or bit, where the first of each is number 1.

예제 입력 1

3
1 0 1 0 1 1 1 0
1 1 0 1 1 1 0 0
1 0 1 1 1 0 0 0
0 0 1 1 1 1 0 1

예제 출력 1

Odd
Byte 3 is broken
Bit 5 is broken

예제 입력 2

4
1 0 1 1 1 0 0 0
1 0 1 1 0 0 0 0
0 1 0 0 1 0 0 0
1 0 1 1 1 0 1 1
1 0 1 1 1 0 1 1

예제 출력 2

Even
Byte 2 is broken
Bit 2 is broken

힌트

Explanation 1

Bytes 1 and 2 have an odd number of 1s but byte 3 has an even number. So odd parity is being used but byte 3 is broken.

The parity byte gives all columns of bits an odd number of 1s except for 5 where they are even, so bit 5 is broken. Bit 5 of byte 3 is corrupt. 

Explanation 2

The first 4 bytes all have an even number of 1s except for byte 2 which is the broken byte.

The parity byte gives all columns of bits an even number of 1s except for 2 where they are odd, so bit 2 is broken. Bit 2 of byte 2 is corrupt.