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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main{  
    public static void main (String[] args){
        FastReader s = new FastReader();
        int t=1;t=s.ni();
        for(int test=1;test<=t;test++){int n,sum=0;
        String S=s.nextLine();n=S.length();
        if(S.contains("0")==false){System.out.println(0);continue;}
        if(S.contains("1")==false){System.out.println(1);continue;}
        for(int i=0;i<n;i++){
            if(S.charAt(i)=='0')sum+=1;
        }
        System.out.println(Math.min(sum,2));
        }
    }
    
    
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
        br = new BufferedReader(new InputStreamReader(System.in));}
        int ni() { return Integer.parseInt(next()); }
        long nl() { return Long.parseLong(next()); }
        double nd() { return Double.parseDouble(next()); }
        int[] readArray(int n){
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] =Integer.parseInt(next());
		return a;
	    }
        String next(){
            while (st == null || !st.hasMoreElements()) {try {st = new StringTokenizer(br.readLine());}
            catch (IOException e){e.printStackTrace();}}return st.nextToken();}
            String nextLine(){String str = "";try {str = br.readLine();}catch (IOException e) 
            {e.printStackTrace();}return str;}
    }
}


  