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.*;

public class CRF16B {

    // Snippet for Fast Input-Output

    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());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static FastReader sc = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

    static int ni() throws IOException {
        return sc.nextInt();
    }

    static long nl() throws IOException {
        return sc.nextLong();
    }

    static int[][] nai2(int n, int m) throws IOException {
        int a[][] = new int[n][m];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                a[i][j] = ni();
        return a;
    }

    static int[] nai(int N, int start) throws IOException {
        int[] A = new int[N + start];
        for (int i = start; i != (N + start); i++) {
            A[i] = ni();
        }
        return A;
    }

    static Integer[] naI(int N, int start) throws IOException {
        Integer[] A = new Integer[N + start];
        for (int i = start; i != (N + start); i++) {
            A[i] = ni();
        }
        return A;
    }

    static long[] nal(int N) throws IOException {
        long[] A = new long[N];
        for (int i = 0; i != N; i++) {
            A[i] = nl();
        }
        return A;
    }

    static void close() {
        out.flush();
    }

    // ---------------------Main-Code------------------------------------------

    public static void main(String[] args) throws Exception {
        int t = ni();
        while (t-- != 0) {
            String str = sc.next();
            int n = str.length();
            if (n == 1 && str.charAt(0) == '1') {
                System.out.println(0);
                continue;
            }
            int count01 = 0;
            boolean isOne = false;
            int firstOne = -1;
            int lastOne = -1;
            for (int i = 0; i < n - 1; i++) {
                char ch1 = str.charAt(i);
                char ch2 = str.charAt(i + 1);
                if (ch1 == '1') {
                    if (firstOne == -1)
                        firstOne = i;
                    isOne = true;
                }
                if (ch2 == '1') {
                    lastOne = i + 1;
                    isOne = true;
                }

                if (ch1 != ch2) {
                    count01++;
                }
            }

            if (count01 == 1 && (firstOne == 0 || lastOne == (n - 1))) {
                System.out.println(1);
                continue;
            }

            if (count01 == 0 && !isOne) {
                System.out.println(1);
                continue;
            }

            if (count01 == 0 && isOne) {
                System.out.println(0);
                continue;
            }

            if (count01 == 2 && (firstOne == 0 && lastOne == (n - 1))) {
                System.out.println(1);
                continue;
            }

            System.out.println(2);
        }
    }

}
