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.
/*==========================================================================
*   	AUTHOR:    RonWonWon
*   	CREATED:   13.09.2021 17:50:07        
/*==========================================================================*/
import java.io.*;
import java.util.*;

public class B {
	
	public static void main(String[] args) throws IOException {
		/*
    in.ini();
    PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
    */
    long startTime = System.nanoTime();
    int t = in.nextInt(), T = 1;  
    while(t-->0) { 
      char ch[] = in.next().toCharArray();
      int c0 = 0, c1 = 0;
      int ans = 0, prev = 0;
      for(char c : ch){
      	if(c=='0'){
      		if(prev==1&&c0!=0){
      			ans = 2; break;
      		}
      		c0++; prev = 0;
      	}
      	else{
      		c1++;
      		prev = 1;
      	}
      }
      if(ans==0){
      	if(c0!=0) ans = 1;
      }
      out.println(ans);
   		//out.println("Case #"+T+": "+ans); T++;
    }
    /*
    startTime = System.nanoTime() - startTime;
    System.err.println("Time: "+(startTime/1000000));
    */
    out.flush();
	}
	
	static FastScanner in = new FastScanner();
  static PrintWriter out = new PrintWriter(System.out);
	static int oo = Integer.MAX_VALUE;
	static long ooo = Long.MAX_VALUE;
	
	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		
		void ini() throws FileNotFoundException {
			br = new BufferedReader(new FileReader("input.txt"));
		}
		
		String next() {
			while(!st.hasMoreTokens())
				try { st = new StringTokenizer(br.readLine()); }
				catch(IOException e) {}
			return st.nextToken();
		}
		
		String nextLine(){
			try{ return br.readLine(); } 
			catch(IOException e) { } return "";
		}
		
		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());
		}
		
		long[] readArrayL(int n) {
			long a[] = new long[n];
			for(int i=0;i<n;i++) a[i] = nextLong();
			return a;
		}
		
		double nextDouble() {
			return Double.parseDouble(next());
		}
		
		double[] readArrayD(int n) {
			double a[] = new double[n];
			for(int i=0;i<n;i++) a[i] = nextDouble();
			return a;
		}
	}

	static final Random random = new Random();

	static void ruffleSort(int[] a){
		int n = a.length;
		for(int i=0;i<n;i++){
			int j = random.nextInt(n), temp = a[j];
			a[j] = a[i]; a[i] = temp;
		}
		Arrays.sort(a); 	
	}
	
	static void ruffleSortL(long[] a){
		int n = a.length;
		for(int i=0;i<n;i++){
			int j = random.nextInt(n);
			long temp = a[j];
			a[j] = a[i]; a[i] = temp;
		}
		Arrays.sort(a); 	
	}
}
