Problem: There is a long blackboard with 2 rows and N columns in a classroom of Kyoto University. This blackboard is so long that it is impossible to tell which cells are already used and which unused.

Recently, a blackboard retrieval device was installed at the classroom. To use this device, you type a search query that forms a rectangle with 2 rows and any length of columns, where each cell is used or unused. When you input a query, the decive answers whether the rectangle that corresponds to the query exists in the blackboard. Here, for a rectangle that corresponds to a search query, if two integer i, j ( i < j ) exist and the rectangle equals to the partial blackboard between column i and j , the rectangle is called a sub-blackboard of the blackboard.

You are currently preparing for a presentation at this classroom. To make the presentation go well, you decided to write a program to detect the status of the whole blackboard using the retrieval device. Since it takes time to use the device, you want to use it as few times as possible.

The status of the whole blackboard is already determined at the beginning and does not change while you are using the device.



Input

The first input is given in the following format:


N


N (1 \leq N \leq 100) is an integer that represents the length of the blackboard.

After this input value, your program must print search queries. A search query has the following format.


s_1
s_2


Here, s_1 represents the upper part of the blackboard and s_2 represents the lower. `#` in s_1 and s_2 represents the cell is already used and `.` represents the cell is still unused. The lengths of s_1 and s_2 are arbitrary, but they must be the same. Make sure to insert a line break at the end of the lines.

Every time your program prints a search query, a string that represents the search result of the device is returned in the followin format.


r


r is either `T` or `F` . The meaning of each character is as follows.

* `T` represents that the sub-blackboard that corresponds to the search query exists in the blackboard.
* `F` represents that the sub-blackboard that corresponds to the search query does not exist in the blackboard.



If the search query equals to the whole blackboard or the number of the search queries exceeds the limit, string `end` is given instead of r . Once you receive this string, exit your program immediately. If your program prints the whole blackboard as a search query before exceedin the limit, it is judged as Accepted. Note that the search query that represents the whole blackboard is also counted as the number of search queries.

Example

Input




Output

Solution (in CPP):

Solution:
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAX(a,b) ((a) > (b) ? (a) : (b))
#define MIN(a,b) ((a) < (b) ? (a) : (b))
#define MEM(a,b) memset((a),(b),sizeof(a))
const LL INF = 1e9 + 7;
const int N = 1e3 + 10;
string x = "..##";
string y = ".#.#";
class AbstactSolver
{
public:
	virtual void outputGuess() = 0;
	virtual void getResult() = 0;
	virtual void outputFinalResult() = 0;
	void solve()
	{
		string s1, s2;
		cin >> n;
		int len = 0;
		int flag = 1;
		char c;
		while (1)
		{
			if (s1.length() == n) break;
			if (flag)
			{
				int key = -1;
				for (int i = 0; i < 4; i++)
				{
					gx = s1 + x[i];
					gy = s2 + y[i];
					outputGuess();
					getResult();
					c = response[0];
					if (c == 'T')
					{
						key = i;
						break;
					}
				}
				if (key == -1) flag = 0;
				else
				{
					s1 = gx;
					s2 = gy;
				}
			}
			else
			{
				int key = -1;
				for (int i = 0; i < 4; i++)
				{
					gx = x[i] + s1;
					gy = y[i] + s2;
					outputGuess();
					getResult();
					c = response[0];
					if (c == 'T')
					{
						key = i;
						break;
					}
				}
				if (key == -1) break;
				s1 = gx;
				s2 = gy;
			}
		}
		// 这里输出最后认为正确的猜测
		gx = s1;
		gy = s2;
		outputFinalResult();
	}
public:
	string gx, gy; // 
	char response[10]; // 反馈的结果
	int n;
};

// LocalSolve用于解决本地调试的情况
class LocalSolver : public AbstactSolver
{
public:
	string cx, cy; // 这里是提供给我们要猜的数字
	int guessCount; // 记录猜测次数
	LL pre;
public:
	LocalSolver()
	{
		guessCount = 0;
		// 这里放入正确的结果，也就是我们要去猜的数
		cx = "#.";
		cy = "#.";
	}
	void outputGuess()
	{
		// 这里之前已经把要猜测的东西保存下来了，只记录猜测次数
		printf("the %d round, you guess the number is: %s %s\n", ++guessCount, gx.c_str(), gy.c_str());
	}


	void getResult()
	{
		// 这里写本地调试器，不要在意细节
		int k = gx.length();
		response[0] = 'F';
		for (int i = 0; i + k <= n; i++)
		{
			if (gx == cx.substr(i, k) && gy == cy.substr(i, k)) response[0] = 'T';
		}
		printf("the %d round, the feedback is:\n%c\n\n", guessCount, response);
	}

	bool isGuessCorrect()
	{
		return gx == cx &&
			gy == cy;
	}
	void outputFinalResult()
	{

		printf("\nYou guess the number is: %s %s\n,", gx.c_str(), gy.c_str());
		printf("The correct Number is: %s %s\n", cx.c_str(), cy.c_str());
		printf("your guess is %s\n", isGuessCorrect() ? "TRUE" : "FALSE");
		printf("total guess count is: %d\n", guessCount);
	}

};

class RemoteSolver : public AbstactSolver
{
public:
	void outputGuess()
	{
		printf("%s\n%s\n", gx.c_str(), gy.c_str());
		fflush(stdout);
	}
	void getResult()
	{
		scanf("%s", response);
	}
	void outputFinalResult()
	{
		printf("%s\n%s\n", gx.c_str(), gy.c_str());
		fflush(stdout);
	}
};
int main()
{
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	//#ifdef ONLINE
	RemoteSolver solver;
	//#else
	//	LocalSolver solver;
	//#endif

	solver.solve();
	return 0;
}