Problem: Fibonacci numbers are the sequence of integers: f0 = 0, f1 = 1, f2 = 1, f3 = 2, f4 = 3, f5 = 5, ..., fn = fn - 2 + fn - 1. So every next number is the sum of the previous two.

Bajtek has developed a nice way to compute Fibonacci numbers on a blackboard. First, he writes a 0. Then, below it, he writes a 1. Then he performs the following two operations:

  * operation "T": replace the top number with the sum of both numbers; 
  * operation "B": replace the bottom number with the sum of both numbers. 



If he performs n operations, starting with "T" and then choosing operations alternately (so that the sequence of operations looks like "TBTBTBTB..."), the last number written will be equal to fn + 1.

Unfortunately, Bajtek sometimes makes mistakes and repeats an operation two or more times in a row. For example, if Bajtek wanted to compute f7, then he would want to do n = 6 operations: "TBTBTB". If he instead performs the sequence of operations "TTTBBT", then he will have made 3 mistakes, and he will incorrectly compute that the seventh Fibonacci number is 10. The number of mistakes in the sequence of operations is the number of neighbouring equal operations («TT» or «BB»).

You are given the number n of operations that Bajtek has made in an attempt to compute fn + 1 and the number r that is the result of his computations (that is last written number). Find the minimum possible number of mistakes that Bajtek must have made and any possible sequence of n operations resulting in r with that number of mistakes.

Assume that Bajtek always correctly starts with operation "T".

Input

The first line contains the integers n and r (1 ≤ n, r ≤ 106).

Output

The first line of the output should contain one number — the minimum possible number of mistakes made by Bajtek. The second line should contain n characters, starting with "T", describing one possible sequence of operations with that number of mistakes. Each character must be either "T" or "B".

If the required sequence doesn't exist, output "IMPOSSIBLE" (without quotes).

Examples

Input

6 10


Output

2
TBBTTB


Input

4 5


Output

0
TBTB


Input

2 1


Output

IMPOSSIBLE

Solution (in JAVA):

Solution:
import java.io.*; 
import java.util.*;
import java.math.*;
import java.awt.geom.*;

import static java.lang.Math.*;

public class Solution implements Runnable {
	
	int n, y;
	
	int getAnswer (int up, int down, int operations, int last) {
		if (up == 0 || down == 0) return Integer.MAX_VALUE / 2;
		if (up == down) {
			if (up == 1 && operations + 1 == n) {
				return last == 0 ? 1 : 0;
			}
			return Integer.MAX_VALUE / 2;
		} else {
			if (up > down) {
				if (down == 1) {
					return up - down - 1 + getAnswer (1, 1, operations + up - down, 0);
				} else {
					return up / down - 1 + getAnswer (up % down, down, operations + up / down, 0);
				}
			} else {
				if (up == 1) {
					return down - up - 1 + getAnswer (1, 1, operations + down - up, 1);
				} else {
					return down / up - 1 + getAnswer (up, down % up, operations + down / up, 1);
				}
			}
		}
	}
	
	String recoverAnswer (int up, int down) {
		StringBuilder sb = new StringBuilder();
		while (true) {
			if (up == down) {
				sb.append("T");
				break;
			}
			if (up > down) {
				up -= down;
				sb.append("T");
			} else {
				down -= up;
				sb.append("B");
			}
		}
		return sb.reverse().toString();
	}
	
	public void solve () throws Exception {
		n = nextInt();
		y = nextInt();
		
		int bestUp = -1, bestDown = -1, bestAnswer = Integer.MAX_VALUE / 2;
		
		for (int x = 1; x <= y; x++) {
			int ans1 = getAnswer (x, y, 0, -1);
			if (ans1 < bestAnswer) {
				bestAnswer = ans1;
				bestUp = x;
				bestDown = y;
			}
			int ans2 = getAnswer (y, x, 0, -1);
			if (ans2 < bestAnswer) {
				bestAnswer = ans2;
				bestUp = y;
				bestDown = x;
			}
		}
		
		if (bestAnswer == Integer.MAX_VALUE / 2) {
			out.println("IMPOSSIBLE");
		} else {
			out.println(bestAnswer);
			out.println(recoverAnswer(bestUp, bestDown));
		}
		
	}
	
	static final String fname = "";
	static long stime = 0;
	
	BufferedReader in;
	PrintWriter out;
	StringTokenizer st;
	
	private String nextToken () throws Exception {
		while (st == null || !st.hasMoreTokens()) {
			st = new StringTokenizer(in.readLine());
		}
		return st.nextToken();
	}
	
	private int nextInt () throws Exception {
		return Integer.parseInt(nextToken());
	}
	
	private long nextLong () throws Exception {
		return Long.parseLong(nextToken());
	}
	
	private double nextDouble () throws Exception {
		return Double.parseDouble(nextToken());
	}
	
	@Override
	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			solve ();
		} catch (Exception e) {
			throw new RuntimeException(e);
		} finally {
			out.close();
		}
	}
	
	public static void main(String[] args) {
		new Thread(null, new Solution(), "", 1<<26).start();
	}

}