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.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;
public class b729 {

	public static void main(String[] args) throws IOException {
			FastReader sc = new FastReader();
	
	int t = sc.nextInt();
	
for(int o = 0 ; o<t;o++) {

	String s = sc.next();
	int n = s.length();
	int c = 0;
	for(int i = 0 ; i<n;i++) {
		if(s.charAt(i) == '1') {
			c++;
		}
	}
	int d = n-c;
	if(c==0) {
		System.out.println(1);
	}else if(c==n) {
		System.out.println(0);
	}else if(c == 1) {
		if(s.charAt(0)=='1' || s.charAt(n-1)=='1') {
			System.out.println(1);
		}else {
			System.out.println(2);
		}
	}else if(d==1) {
		System.out.println(1);
	}
	else {
		int f = 0;
		for(int i = 0 ; i<n;i++) {
			if(s.charAt(i) =='0' ) {
				int j = i;
				int x= 0;
				while(j<n && s.charAt(j) == '0') {
					x++;
					j++;
				}
			if(x == d) {
				f = 1;
			}
			break;
			}
			
		}
		
		if(f == 1) {
			System.out.println(1);
		}
		else
		System.out.println(2);
	}
	
}
	
	}
}


class FastReader {
    BufferedReader br;
    StringTokenizer st;

    public FastReader()
    {
        br = new BufferedReader(
            new InputStreamReader(System.in));
    }

    String next()
    {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());

            
            
            
            
            }
            catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int nextInt() { return Integer.parseInt(next()); }

    long nextLong() { return Long.parseLong(next()); }

    double nextDouble()
    {
        return Double.parseDouble(next());
    }

    String nextLine()
    {
        String str = "";
        try {
            str = br.readLine();
        }
        catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}