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.awt.image.SinglePixelPackedSampleModel;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Arrays;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Queue;
import java.util.Scanner;
public class Main
{
	public static void main(String[] args)
	{
		Scanner in=new Scanner(System.in);
	    int t=in.nextInt();
	    while(t-->0)
	    {
	    	String s=in.next();
	    	char z[]=s.toCharArray();
	    	int sum=0;
	    	for(int i=0;i<s.length();i++)
	    	{
	    		if(z[i]=='0')
	    		{
	    			sum++;
	    			int j=i;
	    			while(j+1<s.length()&&z[j+1]==z[i])
	    			{
	    				j++;
	    			}
	    			i=j;
	    		}
	    	}
	    	if(sum>2)
	    	{
	    		System.out.println(2);
	    	}
	    	else
	    	{
	    		System.out.println(sum);
	    	}
	    }
	}
}
