#include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
using ll = long long;
using ld = long double;
const ll MOD = 1e+9 + 7;
const ll INF = LLONG_MAX;
const int N = (int)2e+5 + 8;
void MAIN(ll tc) {
  int a, b;
  cin >> a >> b;
  if (a < b) {
    if ((a + b) % 2)
      cout << (a + b) / 2 << ' ' << 0 << ' ' << 6 - (a + b) / 2;
    else
      cout << (a + b) / 2 - 1 << ' ' << 1 << ' ' << 6 - (a + b) / 2;
  } else if (a > b) {
    if ((a + b) % 2)
      cout << 6 - (a + b) / 2 << ' ' << 0 << ' ' << (a + b) / 2;
    else
      cout << 6 - (a + b) / 2 << ' ' << 1 << ' ' << (a + b) / 2 - 1;
  } else {
    cout << 0 << ' ' << 6 << ' ' << 0;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed;
  cout << setprecision(10);
  int test_cases = 1;
  for (int i = 1; i <= test_cases; i++) {
    MAIN(i);
  }
}

### Prompt

Please create a solution in Cpp to 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>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
using ll = long long;
using ld = long double;
const ll MOD = 1e+9 + 7;
const ll INF = LLONG_MAX;
const int N = (int)2e+5 + 8;
void MAIN(ll tc) {
  int a, b;
  cin >> a >> b;
  if (a < b) {
    if ((a + b) % 2)
      cout << (a + b) / 2 << ' ' << 0 << ' ' << 6 - (a + b) / 2;
    else
      cout << (a + b) / 2 - 1 << ' ' << 1 << ' ' << 6 - (a + b) / 2;
  } else if (a > b) {
    if ((a + b) % 2)
      cout << 6 - (a + b) / 2 << ' ' << 0 << ' ' << (a + b) / 2;
    else
      cout << 6 - (a + b) / 2 << ' ' << 1 << ' ' << (a + b) / 2 - 1;
  } else {
    cout << 0 << ' ' << 6 << ' ' << 0;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed;
  cout << setprecision(10);
  int test_cases = 1;
  for (int i = 1; i <= test_cases; i++) {
    MAIN(i);
  }
}

```