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.*;
import java.lang.*;

public class C2Q2 {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader( new InputStreamReader(System.in));
        
        int t=Integer.parseInt(br.readLine());
        
        BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
        C2Q2 obj1=new C2Q2();
        while(t-->0){
            String s;
            s=br.readLine();
            output.write(obj1.solve(s)+"\n");
        }
        output.flush();
    }

    public int solve(String s){
        int k=0;
        int zgrps=0;
        int n=s.length();
            while(k<n){
                if(s.charAt(k)=='1'){
                    k++;

                }else{
                    zgrps++;
                    while( (k<n)&&(s.charAt(k)=='0') ){
                        k++;
                    } 
                }
            }
            return  Math.min(2,zgrps);
        }


    }

