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.IOException; 
import java.io.InputStreamReader; 
import java.util.StringTokenizer; 
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.PrintStream;
import java.util.Arrays;
import java.util.HashSet;
public class Sol2 
{ 
	public static  FastReader fs;
	public static void main(String[] args) 
	{ 
		try {
			System.setIn(new FileInputStream("input.txt"));
			System.setOut(new PrintStream(new FileOutputStream("output.txt")));
		} catch (Exception e) {}
		fs = new FastReader();
		int test = fs.nextInt();
		while(test-- > 0){
			solve();
		}
	}

	static 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; 
		} 
	} 
	// use fs as scanner
	// ctrl + shift + b to compile and run 
	public static void solve(){
		// start code here
		
		String str = fs.nextLine();
		int zero = 0, one = 0;
		int sum = 0;
		int len = str.length();
		for(int i=0; i<len; i++){
			char cur = str.charAt(i);
			char pre = i == 0 ? '5' : str.charAt(i-1);
			if(cur == '0'){
				if(cur == pre){
					continue;
				}
				zero++;
			}
			else{
				one++;
			}
		}

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