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.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Random;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
import java.io.*;

	public class A {
		
		BufferedReader br;
		StringTokenizer st;
		
		
		public A(){ // constructor
			
			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());
		}

		String nextLine(){
			
			String str = "";
			try{
				
				str = br.readLine();
			}
			catch (IOException e){
				
				e.printStackTrace();
			}
			return str;
		}
		
		static void sort(int[] arr){
			
	        int n = arr.length;
	        
	        Random rnd = new Random();
	        
	        for(int i = 0; i < n; ++i){
	        	
	            int tmp = arr[i];
	            int randomPos = i + rnd.nextInt(n - i);
	            arr[i] = arr[randomPos];
	            arr[randomPos] = tmp;
	        }
	        
	        Arrays.sort(arr);
	   }
		
		static class Pair{
			int loc;
			int speed;
			Pair(int loc, int speed){
				this.loc = loc;
				this.speed = speed;
			}
		}
		
		static long gcd_long(long a, long b) {
			if(b == 0) {
				return a;
			}
			return gcd_long(b, a % b);
		}
		
		static int mod = (int)1e9 + 7;
		static long mul_long = 1;
		
	public static void main(String[] args) throws Exception{
		
		A in = new A();
		PrintWriter out = new PrintWriter(System.out);
		
		// hard work will power dedication
		
		// Rippling CNIL Spri GOOG.
		
		int test = in.nextInt();
		while(test-- > 0) {
			
			String str = in.next();
			int ans = 0;
			
			for(int i = 0; i < str.length(); i++) {
				char ch = str.charAt(i);
				
				if(ch == '0'){
					ans++;
					while(i < str.length() && str.charAt(i) == '0') {
						i++;
					}
					
				}
			}
			out.println(Math.min(2, ans));
		}
		
		out.flush();
		out.close();
	}
}