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Problem A
UTF-8

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Unicode is an international standard for encoding characters that was designed to overcome the limitations of older text encoding standards like ASCII. The main deficiency of ASCII and similar systems is that they represent only a very limited number of characters (primarily the characters on an English-language keyboard). In contrast, Unicode currently encodes over $130\, 000$ characters (as of 2019), including those from most of the world’s languages, and, in theory, could be used to represent over $1\, 000\, 000$ characters. Perhaps most importantly for modern human communication, Unicode encodes around $3\, 000$ emoji (again, as of 2019).

Unicode has several encoding “flavours,” the most well-known of which are UTF-8, which uses $8$, $16$, $24$, or $32$ bits per character; UTF-16, which uses $16$ or $32$ bits per character; and UTF-32, which uses $32$ bits per character. In all these flavours, the encoding of a character is actually the encoding of a non-negative integer corresponding to the character; this integer is called a code point.1

For this problem, we will focus on UTF-8. A character that is encoded using UTF-8 is stored in $1$, $2$, $3$, or $4$ bytes, and we refer to these four options as Type 1, Type 2, Type 3, and Type 4, respectively. The following table is useful for illustrating these. In each representation of a byte as $8$ bits, the leftmost bit is the most significant.

 

byte $1$

byte $2$

byte $3$

byte $4$

Type 1

0xxxxxxx

     

Type 2

110xxxxx

10xxxxxx

   

Type 3

1110xxxx

10xxxxxx

10xxxxxx

 

Type 4

11110xxx

10xxxxxx

10xxxxxx

10xxxxxx

  • Type 1 – The first (and only) byte begins with 0. The remaining $7$ bits are used to store the code point.

  • Type 2 – The first byte begins with 110, and second byte begins with 10. The remaining $5$ bits of the first byte and the remaining $6$ bits of the second byte ($11$ bits in total) are used to store the code point.

  • Type 3 – The first byte begins with 1110, and the second and third bytes begin with 10. The remaining $4$ bits of the first byte and the remaining $6$ bits of the second and third bytes are used to store the code point ($16$ bits in total).

  • Type 4 – The first byte begins with 11110, and the second, third, and fourth bytes begin with 10. The remaining $3$ bits of the first byte and the remaining $6$ bits of the second, third, and fourth bytes are used to store the code point ($21$ bits in total).

A sequence of bytes adheres to the UTF-8 standard if it consists of one or more character encodings, each of which is of Type 1, Type 2, Type 3, or Type 4, in which case we say the byte sequence is valid UTF-8. Otherwise, the byte sequence is invalid UTF-8. For example, in Sample Input $2$, the first (and only) byte begins with 10, but the first two bits in the first byte of any UTF-8 character encoding can never be 10, so this is invalid UTF-8. And in Sample Input $3$, the first byte should begin a character encoding of Type 4, but only two bytes follow, not three, so this is also invalid UTF-8.

Given a sequence of bytes, determine whether or not it is valid UTF-8, and if it is, report the number of character encodings of Type 1, Type 2, Type 3, and Type 4.

Input

The first line of input contains a positive integer, $n$ ($1 \leq n \leq 1\, 200$), the number of bytes in the sequence. This is followed by $n$ lines giving the $n$ bytes in order, one per line. Each byte is represented as a length-$8$ string of ‘0’ and ‘1’ characters. The leftmost bit of each byte is the most significant bit.

Output

If the sequence of bytes is invalid UTF-8, output a single line containing the word “invalid”. Otherwise output four integers, one per line: the number of character encodings of Type 1, Type 2, Type 3, and Type 4, respectively.

Sample Input 1 Sample Output 1
6
11100011
10001111
10101010
00000000
11011011
10001110
1
1
1
0
Sample Input 2 Sample Output 2
1
10101010
invalid
Sample Input 3 Sample Output 3
3
11110111
10111111
10111111
invalid

Footnotes

  1. Not every code point corresponds to a character, but that is not important here.

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