Problem Description

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

題意：

有兩行數(shù)，初始時(shí)第一行為 0，第二行為 1，現(xiàn)在給出 n、r 兩個(gè)數(shù)，要求對(duì)兩行數(shù)經(jīng)過(guò) n 次操作后得到 r 這個(gè)數(shù)，操作分為兩類：

• T：代表將第二行的數(shù)加到第一行，本次得到的數(shù)是第一行的新數(shù)
• B：代表將第一行的數(shù)加到第二行，本次得到的數(shù)是第二行的新數(shù)

現(xiàn)在要求給出的序列中出現(xiàn)錯(cuò)誤的次數(shù)最小，輸出錯(cuò)誤次數(shù)與操作序列，如果經(jīng)過(guò) n 次操作后達(dá)不到 r，輸出 IMPOSIBLE

思路：

題目本質(zhì)是對(duì)上下兩行數(shù)進(jìn)行操作，得到斐波那契數(shù)列的一項(xiàng)

假設(shè)第一行的數(shù)為 t，第二行的數(shù)為 b，根據(jù)所給的兩個(gè)操作，有：

• 若當(dāng)前執(zhí)行的操作為 T，那么 t=t+b 且 t>b
• 若當(dāng)前執(zhí)行的操作為 B，那么 b=b+t 且 b>t

可以看到，兩個(gè)操作的狀態(tài)是唯一的，那么也就是說(shuō)：

• 若當(dāng)前狀態(tài)為 t>b，則是執(zhí)行操作 T 得來(lái)的
• 若當(dāng)前狀態(tài)為 b>t，則是執(zhí)行操作 B 得來(lái)的

因此可以采用逆推的方式來(lái)得到這個(gè)操作序列，總情況只有 2*r 種，即：枚舉 i，每次通過(guò) (i,r) 與 (r,i) 來(lái)進(jìn)行推導(dǎo)，更新出錯(cuò)的最小值即可

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+7; const int N = 1000000+5; const int dx[] = {-1,1,0,0,-1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std; int n,r; char str[N]; char res[N]; int error=INF; int tot,cnt; int cal(int t,int b){cnt=0;while(t>=0&&b>=0&&t!=b){//逆推if(b>t){str[cnt++]='B';b-=t;}else{str[cnt++]='T';t-=b;}}str[cnt]='T';//首字母為Tif(t!=1||cnt!=n-1)return INF;int minn=0;for(int i=0;i<cnt;i++)if(str[i]==str[i+1])minn++;return minn; } int main() {scanf("%d%d",&n,&r);int error=INF;for(int i=1;i<=r;i++){int minn=cal(i,r);if(minn<error){error=minn;tot=cnt;for(int i=0;i<=cnt;i++)res[i]=str[i];}minn=cal(r,i);if(minn<error){error=minn;tot=cnt;for(int i=0;i<=cnt;i++)res[i]=str[i];}}if(error==INF)printf("IMPOSSIBLE\n");else{printf("%d\n",error);for(int i=tot;i>=0;i--)printf("%c",res[i]);printf("\n");}return 0; }

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