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.util.*;
/*

*/

 
 public class B{
	static FastReader sc=null;
	
	public static void main(String[] args) {
		sc=new FastReader();
		int t=sc.nextInt();
		while(t-->0) {
			char a[]=sc.next().toCharArray();
			int n=a.length;
			int cnt=1;
			for(int i=0;i+1<n;i++)if(a[i]!=a[i+1])cnt++;
			
			if(a[0]=='0')cnt++;
			cnt/=2;
			int ans=Math.min(2,cnt);
			System.out.println(ans);
		}
	}
	
	static int[] ruffleSort(int a[]) {
		ArrayList<Integer> al=new ArrayList<>();
		for(int i:a)al.add(i);
		Collections.sort(al);
		for(int i=0;i<a.length;i++)a[i]=al.get(i);
		return a;
	}
	
	static void print(int a[]) {
		for(int e:a) {
			System.out.print(e+" ");
		}
		System.out.println();
	}
	
	static class FastReader{
		
		StringTokenizer st=new StringTokenizer("");
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		
		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());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		
		int[] readArray(int n) {
			int a[]=new int[n];
			for(int i=0;i<n;i++)a[i]=sc.nextInt();
			return a;
		}
	}
	
	
}
