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.util.ArrayList;




public class Main{
  
    public static void main(String[]args){
        Scanner sc=new Scanner(System.in);
        
    int t=sc.nextInt();
    for (int j=0;j<t;j++){
     String s=sc.next();
     int o=0;
     int l=0;
     for(int i=0;i<s.length();i++){
         if(s.charAt(i)-'0'==0)
         o++;
        
         
         else{
         l++;
             
         }
         
     }
     int oo=0;
     int lo=0;
 for(int i=0;i<s.length();i++){
      if(s.charAt(i)-'0'==0){
          oo++;
      while(i<s.length()&&s.charAt(i)-'0'==0){
      i++;}
      }
  if(i<s.length()&&s.charAt(i)-'0'==1){
      lo++;
        while(i<s.length()&&s.charAt(i)-'0'==1){
      i++;}
      
  }
  i--;
  
      
 }
 
     int ans=0;
     if(o==s.length())
     ans=1;
     if(l==s.length())
     ans=0;
     
     if(oo==1)
     ans=1;
     if(oo>=2)
     ans=2;
     System.out.println(ans);
}
    }
}