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.util.*;
/*
    @stdnoob : 2021-09-17 02:11:47
*/
public class codeforces
{
    public static void main(String args[]) throws Exception
	{
	  	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
        // File f1 = new File("C:\\Users\\Brij\\OneDrive\\Desktop\\Competitive\\input.txt");
        // File f2 = new File("C:\\Users\\Brij\\OneDrive\\Desktop\\Competitive\\output.txt");
        // BufferedReader br = new BufferedReader(new FileReader(f1));
        // PrintWriter pw = new PrintWriter(f2);
        int t = Integer.parseInt(br.readLine());
        while(t-- != 0)
		{
            String s = (br.readLine());
            String ss[] = s.split("2");
            int ans = 0;
            for(int i=0;i<ss.length;i++)
            {
                String ss2[] = ss[i].split("1");
                if(ss2.length!=0)
                {
                    if(ss[i].charAt(0)=='1')
                    ans+=Math.min(ss2.length-1,2);
                    else
                    ans+=Math.min(ss2.length,2);
                }
            }
            pw.println(ans);
		}
		pw.flush();
		pw.close();
		br.close();
    }   
}
