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.util.*;
public class B1566{
	static Scanner sc = null;
	public static void main(String args[]){
		sc = new Scanner(System.in);
		int t = sc.nextInt();
		sc.nextLine();
		while(t-->0){
			String s = sc.nextLine();
			int N = s.length();
			ArrayList<Integer> ar = new ArrayList<>();
			char ch = s.charAt(0);
			ar.add(Integer.parseInt(Character.toString(ch)));
			for(int i=1;i<N;i++){
				char c1 = s.charAt(i);
				if(ch!=c1){
					ar.add(Integer.parseInt(Character.toString(c1)));
				}
				ch = c1;	
			}
			int size = ar.size();
			int count0 = 0;
			int count1 = 0;
			for(Integer j : ar){
				if(j==0)
					count0+=1;
				else
					count1+=1;
			}
			int ans = 0;
			if(size>=4){
				ans = 2;
			}
			else if(size==3){
				if(count1==2)
					ans = 1;
				else
					ans = 2;
			}
			else if(size==2){
				ans = 1;
			}
			else{
				if(count1==1)
					ans = 0;
				else
					ans = 1;
			}
			System.out.println(ans);
		}
	}
}

