A binary string is a string that consists of characters 0 and 1.

Let \operatorname{MEX} of a binary string be the smallest digit among 0, 1, or 2 that does not occur in the string. For example, \operatorname{MEX} of 001011 is 2, because 0 and 1 occur in the string at least once, \operatorname{MEX} of 1111 is 0, because 0 and 2 do not occur in the string and 0 < 2.

A binary string s is given. You should cut it into any number of substrings such that each character is in exactly one substring. It is possible to cut the string into a single substring — the whole string.

A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

What is the minimal sum of \operatorname{MEX} of all substrings pieces can be?

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.

Each test case contains a single binary string s (1 ≤ |s| ≤ 10^5).

It's guaranteed that the sum of lengths of s over all test cases does not exceed 10^5.

Output

For each test case print a single integer — the minimal sum of \operatorname{MEX} of all substrings that it is possible to get by cutting s optimally.

Example

Input


6
01
1111
01100
101
0000
01010


Output


1
0
2
1
1
2

Note

In the first test case the minimal sum is \operatorname{MEX}(0) + \operatorname{MEX}(1) = 1 + 0 = 1.

In the second test case the minimal sum is \operatorname{MEX}(1111) = 0.

In the third test case the minimal sum is \operatorname{MEX}(01100) = 2.
import java.util.*;
import java.io.*;
public class p1
{
    public static void main(String[] args)
    {
        Scanner scan = new Scanner(System.in);
        int tc = scan.nextInt();
        for(int q=0; q<tc; q++)
        {
            String s = scan.next();
            boolean all1 = true;
            boolean all0 = true;
            int count0 = 0;
            for(int i=0; i<s.length(); i++)
            {
                if(s.substring(i,i+1).equals("0"))
                {
                    all1 = false;
                    count0++;
                }
                else
                    all0 = false;
            }
            
            if(all1)
                System.out.println(0);
            else if(all0)
                System.out.println(1);
            else
            {
                System.out.println(Math.min(count0, 2));
            }
        }
    }

}
