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.*;

public class Main {
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer st;
	static PrintWriter pr = new PrintWriter(new OutputStreamWriter(System.out));
	static String readLine() throws IOException {
		return br.readLine();
	}
	static String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(readLine());
		return st.nextToken();
	}
	static int readInt() throws IOException {
		return Integer.parseInt(next());
	}
	static long readLong() throws IOException {
		return Long.parseLong(next());
	}
	static double readDouble() throws IOException {
		return Double.parseDouble(next());
	}
	static char readChar() throws IOException {
		return next().charAt(0);
	}
	static class Pair implements Comparable<Pair> {
		int f, s;
		Pair(int f, int s) {
			this.f = f; this.s = s;
		}
		public int compareTo(Pair other) {
			if (this.f != other.f) return this.f - other.f;
			return this.s - other.s;
		}
	}
	static void solve() throws IOException {
		String s = " " + readLine();
		int n = s.length() - 1, cnt = 0;
		for (int i = 1; i <= n; ++i)
			if (s.charAt(i) == '0') ++cnt;
		if (cnt == 0) pr.println(0);
		else if (cnt == 1 || cnt == n) pr.println(1);
		else pr.println(2);
	}
	public static void main(String[] args) throws IOException {
		//solve();
		for (int t = readInt(); t > 0; --t) solve();
		pr.close();
	}
}
