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.
//package CodeforcesContests;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class B {
    
    public static void main (String[] Z) throws IOException {
        
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder op = new StringBuilder();
        StringTokenizer stz;
        
        int T = Integer.parseInt(br.readLine());
        
        
        while(T-- > 0) {
            
            String s = br.readLine();
            int n = s.length();
            
            int segs = 0;
            
            for(int i = 0 ; i < n ; ++i) {
                
                if(s.charAt(i) == '0') {
                    while (i < n && s.charAt(i) == '0') {
                        ++i;
                        
                    }
                    ++segs;
                }
                
            }
            
            if(segs > 2) {
                op.append("2\n");
            }
            else {
                op.append(segs + "\n");
            }

            
            
            
            
            
        }
        
        System.out.println(op);
        
        
        // END OF CODE
    }
    
}
