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.*;

	public class practice{
		public static void main(String[] args){
			FastScanner r = new FastScanner();					
			int t = r.i() ,n;
			StringBuffer ans = new StringBuffer("");
			char[] ar;
			int min;
			int i;
			while(t-->0) {
				ar =r.n().toCharArray();
				min= ar[0]=='0' ? 1 : 0;
				
				for(i=1;i<ar.length;i++) {
					
					if(ar[i] != ar[i-1]) {
						
						min+=  ar[i]=='0' ? 1 : 0;
					}
				}
				ans.append( Math.min(min, 2) + "\n");
			}				
			p(ans);
		}

		
		static <T> void p(T s) {
			System.out.println(s);
		}
		
		static void p() {
			System.out.println();
		}		
		static <T> void pn(T s) {
			System.out.print(s);
		}

	}
		

	
	class FastScanner {
	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    StringTokenizer st=new StringTokenizer("");
	    String n() {
	        while (!st.hasMoreTokens())
	            try {
	                st=new StringTokenizer(br.readLine());
	            } catch (IOException e) {
	                e.printStackTrace();
	            }
	        return st.nextToken();
	    }
	    int[] ia(int n) {
	    	int[] arr = new int[n];
	    	for(int i=0;i<n;i++)
	    		arr[i] = i();
	    	return arr;
	    }
	    String[] sa(int n) {
	    	String[] arr = new String[n];
	    	for(int i=0;i<n;i++)
	    		arr[i] = n();
	    	return arr;
	    }
	    long[] la(int n) {
	    	long[] arr = new long[n];
	    	for(int i=0;i<n;i++)
	    		arr[i] = l();
	    	return arr;
	    }
	    String nl()
	    {
	        String str = "";
	        try
	        {
	            str = br.readLine();
	        }
	        catch (IOException e)
	        {
	            e.printStackTrace();
	        }
	        return str;
	    }

	    int i() {
	        return Integer.parseInt(n());
	    }
	    long l() {
	        return Long.parseLong(n());
	    }
	}
