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.math.*;
import java.lang.Math;

public final class B_MIN_MEX_Cut {
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));

        // Start writing your solution here.
        int testCases = sc.nextInt();
        for (int cases = 0; cases < testCases; cases++) {
            // My code
            String binary = sc.next();
            boolean zero = binary.contains("0");
            boolean one = binary.contains("1");

            // checking if all zeros or all ones;
            if (!one || !zero) {
                if (one) {
                    // all ones
                    out.println(0);
                } else {
                    // all zeros
                    out.println(1);
                }

                // not all ones or zeros
            } else if (binary.length() <= 3) {
                // out.println("TBD");
                // if count of zero is more => 2 ( ex 010 , 001)
                // if count of one is more => 1 (ex 101 , 110 , 011)
                // if count equal => 1
                int c0 = 0;
                for (int i = 0; i < binary.length(); i++) {
                    if (binary.charAt(i) == '0') {
                        c0++;
                    }
                }
                int c1 = binary.length() - c0;
                if (c1 == c0) {
                    out.println(1);
                } else if (c1 > c0) {
                    out.println(1);
                } else if (c0 > c1) {
                    out.println(2);
                }
            } else {
                if (zero && one) {
                    // if count of zero is only 1, rest are ones
                    int c0 = 0;
                    int c1 = 0;
                    for (int i = 0; i < binary.length(); i++) {
                        if (binary.charAt(i) == '0') {
                            c0++;
                        }
                        if (binary.charAt(i) == '1') {
                            c1++;
                        }
                    }
                    if (c0 == 1) {
                        out.println(1);
                    } else if (c0 > 1 && c1 == 1) {
                        if (binary.charAt(0) == '1' || binary.charAt(binary.length() - 1) == '1') {
                            // one 1 at extremes , then mex = 1
                            out.println(1);
                        } else {
                            out.println(2);
                        }
                    } else {
                        out.println(2);
                    }
                } else if (zero) {
                    // only zeros
                    out.println(1);
                } else {
                    // only 1s
                    out.println(0);
                }
            }

        }
        // Stop writing your solution here.
        out.close();
    }

    // -----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;

    // -----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            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;
        }
    }
}