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 {
	public static void main(String args[]){
		FScanner in = new FScanner();
		PrintWriter out = new PrintWriter(System.out);
		int t = in.nextInt();
		while(t-->0) { 
			String s=in.next();
			int c0=0,c1=0,c2=0;

			for(int i=0;i<s.length();i++)
			{
				char ch=s.charAt(i);
				int val=ch-'0';
				if(ch=='0')
				{
					c0++;
				}
				else if(ch=='1')
					c1++;
				
			}

			if(c0==0)
				out.println(0);
			else if(c0>=2&&c1>=1)
				out.println(2);
			else
				out.println(1);
		}
		out.close();
	}

	static boolean checkprime(int n1)
	{
		if(n1%2==0||n1%3==0)
			return false;
		else 
		{
			for(int i=5;i*i<=n1;i+=6)
			{
				if(n1%i==0||n1%(i+2)==0)
					return false;
			}
			return true;
		}
	}
	
	static class FScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer sb = new StringTokenizer("");

		String next(){
			while(!sb.hasMoreTokens()){
				try{
					sb = new StringTokenizer(br.readLine());
				} catch(IOException e){ }
			}
			return sb.nextToken();
		}
		String nextLine(){
			try{ return br.readLine(); } 
			catch(IOException e) { } return "";
		}

		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] = nextInt();
			return a;
		}

		float nextFloat(){
		return Float.parseFloat(next());
		}
		
		double nextDouble(){
		return Double.parseDouble(next());
		}
	}
}
