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.InputStreamReader;
import java.util.HashSet;

public class Main {
    public static void main(String[] args) throws java.lang.Exception {
        InputStreamReader in = new InputStreamReader(System.in);
        BufferedReader br = new BufferedReader(in);
        String s = br.readLine();
        long t = Long.parseLong(s);
//        int r=1;
        while (t-->0) {
             s = br.readLine();
//             String ss[] = s.trim().split(" ");
            int zerogrp = 0;
            int ones=0;
            int zero=0;
            for(int i=0;i<s.length();i++){
                if(s.charAt(i)=='0'){
                    zero++;
                    if(i<s.length()-1){
                        if(s.charAt(i+1)=='1'){
                            zerogrp++;
                        }
                    }
                }else{
                    ones++;
                }
            }
            if(s.charAt(s.length()-1)=='0'){
                zerogrp++;
            }

            if(ones==0 && zero>0){
                System.out.println(1);
            }else if(zero==0 && ones>0){
                System.out.println(0);
            }else if(zerogrp<=2){
                System.out.println(zerogrp);
            }else {
                System.out.println(2);
            }


        }
    }
}

