Codeforce Round #222 Div2 A

A. Playing with Dice

time limit per test 1 second

memory limit per test 256 megabytes

 

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|.

#pragma comment(linker,"/STACK:102400000,102400000")
#include <cstdio>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x7fffffff
#define mod 1000000007
#define ll long long
#define maxn 506
#define pi acos(-1.0)                                             
#define FF(i,n) for(int i=0;i<n;i++)
int n, m, z, t, flag, k;
int main(){
    cin >> n >> m;
    int nn = max(n, m), mm = min(n, m);
    z = 6 - (nn + mm) / 2;
    if (abs(n - m) % 2 == 0)t = 1;
    if (n == m)t = 6,z=0,k=0;
    k = 6 - t - z;
    if (n<m)swap(z, k);
    printf("%d %d %d\n", z, t, k);
}