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.*;
import java.io.*;
public class B{
	public static void main(String[] args)
	{
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int t = fs.nextInt();
		for(int tt=0;tt<t;tt++){
		char[] arr = fs.next().toCharArray();
		int min = 2;
		
		int one = 0; int zero = 0;
		
		for(char ch:arr)
		{
		    if(ch == '1')one++;
		    else zero++;
		}
		
		if(zero == 0)
		{
		    min = 0;
		}
		else if(one == 0)
		{
		    min = 1;
		}
		else{
		int count = 0;
		
		if(arr[0] == '0')count++;
		
		for(int i=1;i<arr.length;i++)
		{
		    if(arr[i] == '0' && arr[i-1] != '0')count++;
		}
		
		if(count<2)min = 1;
		}
		
		out.println(min);
		}
		
		out.close();		
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	public static int[] sort(int[] arr)
	{
		List<Integer> temp = new ArrayList();
		for(int i:arr)temp.add(i);
		Collections.sort(temp);
		int start = 0;
		for(int i:temp)arr[start++]=i;
		return arr;
	}
	public static String rev(String str)
	{
		char[] arr = str.toCharArray();
		char[] ret = new char[arr.length];
		int start = arr.length-1;
		for(char i:arr)ret[start--] = i;
		String ret1 = "";
		for(char ch:ret)ret1+=ch;
		return ret1;
	}
}