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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Scanner;

public class Codeforcesmaytwentyone {
    public static void main(String[] args) throws IOException {
        try {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
            
             int t = Integer.parseInt(br.readLine().trim());
             while(t-->0){
                String s = br.readLine(); 
                char arr[] = s.toCharArray();
                long sum =0;
                long count1=0;
                long count0=0;
                for(int i=0; i<s.length(); i++){
                  if(arr[i]=='0'){sum+=1; count0++;}
                  else{sum+=0; count1++;}
                }
               if(count0==0){
                   sum=0;
               }
               else if(count1==0){
                   sum=1;
               }
               else if(count1==1||count0==1){
                   sum=1;
               }
                System.out.println(sum);
                  
             }
            br.close();
            log.flush();
        } catch(Exception e) {
            return;
        }
    }   // end-main

    static void print(int arr[]) {
        System.out.print("Print :- ");
        for(int i=0 ; i<arr.length ; i++) {
            System.out.print(arr[i]+" ");
        }
        System.out.println();
    }   // end-print
    static boolean isPrime(int n)
    {
 
        // Check if number is less than
        // equal to 1
        if (n <= 1)
            return false;
 
        // Check if number is 2
        else if (n == 2)
            return true;
 
        // Check if n is a multiple of 2
        else if (n % 2 == 0)
            return false;
 
        // If not, then just check the odds
        for (int i = 3; i <= Math.sqrt(n); i += 2)
        {
            if (n % i == 0)
                return false;
        }
        return true;
    }
 
 
}
