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.*;
import java.math.BigInteger;

public class P04 {
	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());
		}
		
		float nextFloat() {
			return Float.parseFloat(next());
		}
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			}catch(IOException e) {
				e.printStackTrace();
			}
			return str ;
		}
	}
	static void swap(int[] arr, int i, int j)
	{
	    int temp = arr[i];
	    arr[i] = arr[j];
	    arr[j] = temp;
	}

	static int partition(int[] arr, int low, int high)
	{	   
	    long pivot = arr[high];   
	    int i = (low - 1);
	    for(int j = low; j <= high - 1; j++)
	    {    
	        if (arr[j] < pivot)
	        {          
	            i++;
	            swap(arr, i, j);
	        }
	    }
	    swap(arr, i + 1, high);
	    return (i + 1);
	}
	static void quickSort(int[] arr, int low, int high)
	{
	    if (low < high)
	    {  
	        int pi = partition(arr, low, high);
	        quickSort(arr, low, pi - 1);
	        quickSort(arr, pi + 1, high);
	    }
	}
	public static long binarySearch(long arr[], long key){  
		  int  first =0,last =arr.length-1;
		   int mid = (first + last)/2;  
		   while( first <= last ){  
		      if ( arr[mid] < key ){  
		        first = mid + 1;     
		      }else if ( arr[mid] == key ){  
		    	 return mid;  
		      }else{  
		         last = mid - 1;  
		      }  
		      mid = (first + last)/2;  
		   }  
		   if ( first > last ){ 
			   return -1;
		       
		   } 
		   return -1;
		 }  
	static long fp(long a , long b , int n) {
		long res =1;
		while(b>0) {
			if((b&1)!=0) {             //when b is odd 
				res=(res*a%n)%n;
			}
			a=(a%n*a%n)%n;
			b=b>>1;
		}
		return res;
	}
	static long gcd (long a ,long b) {
		if(b==0) {
			return a;
		}
		return gcd(b,a%b);
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner (System.in);
		FastReader fs = new FastReader();
		int t = fs.nextInt();
		while(t-->0) {
			
			String s = fs.next();
			int n = s.length();
			int ans = 0,c=0;
			for(int i =0;i<n;i++) {
				if(i<n-1) {
					if(s.charAt(i)=='1'&&s.charAt(i+1)=='0'||
							s.charAt(i)=='0'&&s.charAt(i+1)=='1')c++;
				}
				if(s.charAt(i)=='1')ans+=0;
				else ans+=1;
			}
		//	System.out.println(c+" "+ans);
			if(c==1)System.out.println(1);
			else if(ans==n)System.out.println(1);
			else if(ans>2)System.out.println(2);
			else System.out.println(ans);
		}
		
    }
	

	private static int solve(int[] a,int b[], int n) {
		int flag =0;
		for(int j =0;j<n;j++) {
			if(a[j]!=b[j]) {
				flag=1;
			}
		}
		if(flag==0)return 0;
		int i=1;
		while(true) {
			if(i%2==0) {
				for(int j=1;j<n-1;j+=2) {
					if(a[j]>a[j+1]) {
						int temp =a[j];
						a[j]=a[j+1];
						a[j+1]=temp;
					}
				}
			}
			else {
				for(int j =0;j<n-2;j+=2) {
					if(a[j]>a[j+1]) {
						int temp =a[j];
						a[j]=a[j+1];
						a[j+1]=temp;
					}
				}
			}
			flag =0;
			for(int j =0;j<n;j++) {
				if(a[j]!=b[j])flag=1;
			}
			if(flag==0)return i;
			i++;
		}
       
		
		
	}

	private static long sumofn(long n) {
		int ans = 0;
		while(n>9) {
			ans+=n%10;
			n/=10;
		}
		ans+=n;
		return ans;
	}
	private static boolean palin(String s) {
		int i =0,j=s.length()-1;
		while(i<j) {
			if(s.charAt(i)!=s.charAt(j))return false;
			else {
				i++;
				j--;
			}
		}
		return true;
	}
}
			


