#include <bits/stdc++.h>
int main() {
  int a, b, x = 0, y = 0, d = 6;
  scanf("%d %d", &a, &b);
  if (6 - a == b - 1 || a - 1 == 6 - b) {
    x = 3;
    y = 3;
    d = 0;
  } else if (a > b) {
    x = 6 - a + (a - b) / 2 + (a - b) % 2;
    y = b - 1 + (a - b) / 2 + (a - b) % 2;
    d = 6 - (x + y);
  } else if (b > a) {
    x = a - 1 + (b - a) / 2 + (b - a) % 2;
    y = 6 - b + (b - a) / 2 + (b - a) % 2;
    d = 6 - (x + y);
  }
  printf("%d %d %d", x, d, y);
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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.

Examples

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|.

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  int a, b, x = 0, y = 0, d = 6;
  scanf("%d %d", &a, &b);
  if (6 - a == b - 1 || a - 1 == 6 - b) {
    x = 3;
    y = 3;
    d = 0;
  } else if (a > b) {
    x = 6 - a + (a - b) / 2 + (a - b) % 2;
    y = b - 1 + (a - b) / 2 + (a - b) % 2;
    d = 6 - (x + y);
  } else if (b > a) {
    x = a - 1 + (b - a) / 2 + (b - a) % 2;
    y = 6 - b + (b - a) / 2 + (b - a) % 2;
    d = 6 - (x + y);
  }
  printf("%d %d %d", x, d, y);
  return 0;
}

```