For a positive integer n

* If n is even, divide by 2.
* If n is odd, multiply by 3 and add 1.



If you repeat the above operation, the result will be 1. A problem called "Colatz conjecture" is that repeating this operation for any positive integer n will always result in 1. This problem is an unsolved problem, also known as the "Kakutani problem" in Japan. It is known that there is no counterexample for a very large number 3 × 253 = 27,021,597,764,222,976 using a computer, but it has not been mathematically proven.

Create a program that takes the integer n as an input and outputs the number of operations that are repeated until the result is 1. The integer n should be an integer that is 1 or more and the value in the middle of repeating the above calculation is 1000000 or less. For example, if you receive 3 as input, the operation column will be


3 → 10 → 5 → 16 → 8 → 4 → 2 → 1


Therefore, 7 is output, which is the number of operations (the number of arrows above).



Input

A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. One integer n (n ≤ 1000000) is given on one row for each dataset.

The number of datasets does not exceed 50.

Output

Outputs the number of operations for each dataset on one line.

Example

Input

3
10
0


Output

7
6

import java.awt.geom.Point2D;
import java.io.*;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.NoSuchElementException;
import java.util.PriorityQueue;
import java.util.Scanner;

import javax.xml.crypto.Data;
public class Main {

	static PrintWriter out = new PrintWriter(System.out);
	static FastScanner sc = new FastScanner();
	static Scanner stdIn = new Scanner(System.in);

	static int[] dp = new int [10000001];
	public static void main(String[] args) {
		dp();
		while(true) {
			int n = sc.nextInt();
			if(n == 0) break;
			out.println(dp[n]);
		}
		out.flush();
	}
	
	public static void dp() {
		dp[1] = 0;
		for(int i = 2; i < 10000001; i++) {
			solv(i,0,i);
		}
	}
	public static void solv(int a, int b, int c) {
		if(a >= 1000001) {
			dp[c] = -1;
			return;
		}
		if(dp[a] < 0) {
			dp[c] = -1;
			return;
		}
		if(dp[a] >= 1) {
			dp[c] = dp[a] + b;
			return;
		}
		if(a == 1) {
			dp[c] = b;
			return;
		}
		if(a % 2 == 0) {
			solv(a/2,b+1,c);
			return;
		}
		else {
			solv(a*3+1,b+1,c);
			return;
		}
	}
}
		
//------------------------------//
//-----------//
class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        }else{
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }
    private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
    private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
    private void skipUnprintable() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;}
    public boolean hasNext() { skipUnprintable(); return hasNextByte();}
    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while(true){
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            }else if(b == -1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    
    public int nextInt() {
    	return (int)nextLong();
    }
    
    public double nextDouble() {
    	return Double.parseDouble(next());
    }
    

}