Students in a class are making towers of blocks. Each student makes a (non-zero) tower by stacking pieces lengthwise on top of each other. n of the students use pieces made of two blocks and m of the students use pieces made of three blocks.

The students don’t want to use too many blocks, but they also want to be unique, so no two students’ towers may contain the same number of blocks. Find the minimum height necessary for the tallest of the students' towers.

Input

The first line of the input contains two space-separated integers n and m (0 ≤ n, m ≤ 1 000 000, n + m > 0) — the number of students using two-block pieces and the number of students using three-block pieces, respectively.

Output

Print a single integer, denoting the minimum possible height of the tallest tower.

Examples

Input

1 3


Output

9


Input

3 2


Output

8


Input

5 0


Output

10

Note

In the first case, the student using two-block pieces can make a tower of height 4, and the students using three-block pieces can make towers of height 3, 6, and 9 blocks. The tallest tower has a height of 9 blocks.

In the second case, the students can make towers of heights 2, 4, and 8 with two-block pieces and towers of heights 3 and 6 with three-block pieces, for a maximum height of 8 blocks.
#include <bits/stdc++.h>
using namespace std;
void swap(long long int &a, long long int &b) {
  auto tm = a;
  a = b;
  b = tm;
}
long long int power(long long int x, long long int y) {
  long long int sum = 1;
  for (long long int i = 0; i < y; i++) sum *= x;
  return sum;
}
long long int fact(long long int n) {
  long long int res = 1;
  for (long long int i = 2; i <= n; i++) res = res * i;
  return res;
}
long long int nCr(long long int n, long long int r) {
  return fact(n) / (fact(r) * fact(n - r));
}
vector<long long int> adj[200005];
bool vis[200005] = {false};
long long int s[200005] = {0};
long long int k, a[200005] = {0};
long long int ans;
void bfs(long long int node) {
  vis[node] = true;
  for (auto &(i) : adj[node]) {
    if (!vis[i]) {
      bfs(i);
      s[node] += s[i];
      if (s[node] >= k && a[node] == 1) ans = node;
    }
  }
}
void solve() {
  long long int n, i, j;
  long long int m, x, y, z;
  cin >> n >> m;
  for (i = 1;; i++) {
    x = i / 2;
    y = i / 3;
    z = i / 6;
    if ((x + y - z) >= (n + m) && x >= n && y >= m) {
      cout << i;
      return;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int tt = 1.0 * clock();
  int t;
  t = 1;
  for (int y = 0; y < t; y++) {
    solve();
    cout << endl;
  }
}
