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.
// "static void main" must be defined in a public class.
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner  s=  new Scanner(System.in);
        int t=s.nextInt();
        for(int i=0;i<t;i++)
        {
           String str = s.next();
            
            System.out.println(fn(str));
        }
    }
    
    public static int fn(String str)
    {
    int nz=0;
        int no=0;
        int nt=0;
        int fz=-1;
        int lz=-1;
        int fo=-1;
        int lo=-1;
        int i=0;
        for(char c:str.toCharArray())
        {
            if(c=='0')
               { nz++;  
                
              
                }
            else if(c=='1')
               { no++; 
                
                }
            else
                nt++;
            
            i++;
        }
        
        if(nz==0)
            return 0;
        
        if(no==0)
            return 1;
        
       int in=0;
        int tot=0;
        while(in<str.length())
        {
            char m  = str.charAt(in);
            in++;
            while(in<str.length() && str.charAt(in)==m)
                m++;
           if(m=='0')
               tot+=1;
        
        }
        
        return Math.min(2,tot);
        
        
    }
}