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.lang.*;
import java.io.*;
 
public class main
{
    
    
    
	public static void main (String[] args) throws java.lang.Exception
    {
		FastReader sc=new FastReader();
            int t = sc.nextInt();
            while(t-->0){
                String s = sc.next();
                
                int n = s.length();
                int i = 0;
                int mex = 0;
                while(i<n){
                    while(i<n && s.charAt(i)=='0'){
                        i++;
                        mex++;
                    }
                    while(i<n && s.charAt(i) == '1'){
                        i++;
                    }
                }
                
                System.out.println(Math.min(2,mex));
            }  
    }
}

class pair{
    int v;
    int i;
    pair(int v , int i){
        this.v = v;
        this.i = i;
    }
}
 
class FastReader {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader() {br = new BufferedReader(new InputStreamReader(System.in));}
 
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();}
 
        int nextInt() { return Integer.parseInt(next());}
        long nextLong() { return Long.parseLong(next());}
        double nextDouble() { return Double.parseDouble(next());}
        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;}
}