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.*;
public class class251 {
public static void main(String arg[])
{
     Scanner sc=new Scanner(System.in);
     int t=sc.nextInt();
     while(t-->0)
     {
    	 String s=sc.next();
    	 int l=s.length();
    	 char ch[]=s.toCharArray();
    	 int i,c=0,flag=0,j;
    	 for(i=0;i<l;i++)
    	 {
    		 if(ch[i]=='0')
    		 {
    			 c++;
    			 for(j=i+1;j<l;j++)
    			 {
    				 if(ch[j]=='1')
    				 {
    					 i=j;
    					 break;
    				 }
    			 }
    		 }
    	 }
    	 int ans=Math.min(2,c);
    	 System.out.println(ans);
    	 
     }
}
}
