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.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class JaiShreeRam{
	static Scanner in=new Scanner();
	static long mod = 1000000007;
	static ArrayList<ArrayList<Integer>> adj;
	public static void main(String[] args) throws Exception{
		int z=in.readInt();
		while(z-->0){
			solve();
		}
	}
		
	static void solve() {
		char c[]=in.readString().toCharArray();
		int n=c.length;
		Set<Character> st=new HashSet<>();
		for(int i=0;i<n;i++) {
			st.add(c[i]);
		}
		if(!st.contains('0')) {
			System.out.println(0);
		}
		else if(!st.contains('1')){
			System.out.println(1);
		}
		else {
			int ans=2;
			int cnt=0;
			int k=0;
			for(int i=0;i<n;i++) {
				if(c[i]=='0') {
					cnt++;
				}
				else {
					if(cnt>0) {
						k++;
						cnt=0;
					}
				}
			}
			if(cnt>0) {
				k++;
				cnt=0;
			}
			System.out.println(Math.min(k, ans));
		}
	}
	
	static int[] nextIntArray(int n){
		int[] arr= new int[n];
		int i=0;
		while(i<n){
			arr[i++]=in.readInt();
		}
		return arr;
	}
	static int[] nextIntArray1(int n){
		int[] arr= new int[n+1];
		int i=1;
		while(i<=n){
			arr[i++]=in.readInt();
		}
		return arr;
	}
	static long gcd(long a, long b) {
		if (b==0) return a;
		return gcd(b, a%b);
	}
	static class Scanner{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String readString() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		double readDouble() {
			return Double.parseDouble(readString());
		}
		int readInt() {
			return Integer.parseInt(readString());
		}
		long readLong() {
			return Long.parseLong(readString());
		}
	}	
}