codeforce

A. Playing with Dice

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Two players are playing a game. First each of them writes an integer from 1 to 6, and then a dice is thrown. The player whose written number got closer to the number on the dice wins. If both payers have the same difference, it's a draw.

The first player wrote number a, the second player wrote number b. How many ways to throw a dice are there, at which the first player wins, or there is a draw, or the second player wins?

Input

The single line contains two integers a and b (1 ≤ a, b ≤ 6) — the numbers written on the paper by the first and second player, correspondingly.

Output

Print three integers: the number of ways to throw the dice at which the first player wins, the game ends with a draw or the second player wins, correspondingly.

Sample test(s)

input

2 5

output

3 0 3

input

2 4

output

2 1 3

Note

The dice is a standard cube-shaped six-sided object with each side containing a number from 1 to 6, and where all numbers on all sides are distinct.

You can assume that number a is closer to number x than number b, if |a - x| < |b - x|.

做了一题水题，学校太坑，晚上想做场比赛都费劲，一直断电，一会睡觉吧。

 #include<stdio.h>
 int main(int argc, char const *argv[])
 {
     int a, b;
     while(~scanf("%d%d", &a, &b))
     {
         if(!((a + b) % ))
         {
             if(a == b)
             {
                 printf("0 6 0\n");
             }
             else if(a < b)
             {
                 printf("%d ", (a + b)/ - );
                 printf("1 ");
                 printf("%d\n",  - (a+b)/);
             }
             else
             {
                 printf("%d ", -(a+b)/);
                 printf("1 ");
                 printf("%d\n", (a+b)/ - );
             }
         }
         else
         {
             if(a < b)
             {
                 printf("%d ", (a + b)/);
                 printf("0 ");
                 printf("%d\n",  - (a + b)/);
             }
             else
             {
                 printf("%d ", -(a + b)/);
                 printf("0 ");
                 printf("%d\n", (a+b)/);
             }
         }
     }
     return ;
 }