#include <bits/stdc++.h>
using namespace std;
template <class TH>
void _dbg(const char* sdbg, TH h) {
  cerr << sdbg << " = " << h << endl;
}
template <class TH, class... TA>
void _dbg(const char* sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << " = " << h << ", ";
  _dbg(sdbg + 1, a...);
}
const int maxn = (1e6) + 7;
const int maxk = 20;
const int inf = (1e9) + 7;
const long long LLinf = ((long long)1e18) + 7LL;
const long double eps = 1e-9;
const long long mod = 1e9 + 7;
struct Matching {
  int n, wynik;
  vector<int> odl, str, para, odw;
  const vector<vector<int> >& V;
  Matching(int N, const vector<vector<int> >& M, const vector<int>& s)
      : n(N), odl(n + 1), str(s), para(n + 1), odw(n + 1), V(M) {}
  bool DFS(int v, int ojciec, int zostalo) {
    if (zostalo < 0) return false;
    odw[v] = 1;
    if (!str[v]) {
      if (!para[v]) return (para[v] = ojciec);
      if (DFS(para[v], v, zostalo - 1)) return (para[v] = ojciec);
      return false;
    }
    for (int i = 0; i < ((int)(V[v]).size()); i++)
      if (odl[V[v][i]] == odl[v] + 1 && !odw[V[v][i]] &&
          DFS(V[v][i], v, zostalo - 1))
        return (para[v] = V[v][i]);
    return false;
  }
  int sciezki() {
    memset(&odl[0], 0, 4 * (n + 1));
    memset(&odw[0], 0, 4 * (n + 1));
    int minodl = 2 * n + 5, ilezmian = 0;
    queue<pair<int, int> > Q;
    for (int i = 1; i <= n; i++)
      if (str[i] && !para[i]) {
        Q.push(make_pair(i, 1));
        odl[i] = 1;
      }
    while (!Q.empty()) {
      int akt = Q.front().first, od = Q.front().second;
      Q.pop();
      if (!str[akt]) {
        if (para[akt]) {
          Q.push(make_pair(para[akt], od + 1));
          odl[para[akt]] = od + 1;
        } else
          minodl = min(minodl, od);
        continue;
      }
      for (int i = 0; i < ((int)(V[akt]).size()); i++)
        if (odl[V[akt][i]] == 0) {
          Q.push(make_pair(V[akt][i], od + 1));
          odl[V[akt][i]] = od + 1;
        }
    }
    for (int i = 1; i <= n; i++)
      if (!odw[i] && str[i] && !para[i]) ilezmian += DFS(i, -1, minodl - 1);
    return ilezmian;
  }
  int matching() {
    int ile = 0, akt;
    while ((akt = sciezki())) ile += akt;
    return wynik = ile;
  }
  vector<int> mpn() {
    vector<int> stan(n + 1, 0);
    queue<int> Q;
    for (int i = 1; i <= n; i++) {
      if (!para[i]) {
        stan[i] = 1;
        Q.push(i);
      }
    }
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (auto el : V[x]) {
        if (stan[el] == 0) {
          stan[el] = -1;
          stan[para[el]] = 1;
          Q.push(para[el]);
        }
      }
    }
    vector<int> odp;
    for (int i = 1; i <= n; i++)
      if (stan[i] == 1 || (!stan[i] && str[i])) odp.push_back(i);
    return odp;
  }
};
vector<pair<int, int> > mov = {{-1, -2}, {-1, 2}, {-2, -1}, {-2, 1},
                               {1, -2},  {1, 2},  {2, -1},  {2, 1},
                               {3, 0},   {0, 3},  {-3, 0},  {0, -3}};
int A, B;
int numer(int a, int b) { return (a - 1) * B + b; }
long long wyn(int a, int b) {
  A = a;
  B = b;
  vector<vector<int> > V;
  vector<int> cnt;
  V.resize(numer(a, b) + 10);
  cnt.resize(numer(a, b) + 10);
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++) {
      for (auto s : mov) {
        int ii = i + s.first;
        int jj = j + s.second;
        if (1 <= ii && ii <= a && 1 <= jj && jj <= b) {
          V[numer(i, j)].push_back(numer(ii, jj));
        }
      }
      cnt[numer(i, j)] = (i + j) & 1;
    }
  Matching xd(numer(a, b) + 5, V, cnt);
  return xd.matching() * 2LL;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long x, y;
  cin >> x >> y;
  long long res = 0LL;
  if (x > y) swap(x, y);
  if (x == 1LL) {
    long long cnt = y / 6;
    y -= cnt * 6;
    cout << cnt * 6 + max(0LL, y - 3) * 2LL;
    return 0;
  }
  while (x * y > 1000) {
    if (x > 20) {
      if (x > 20050) {
        res += 20000LL * y;
        x -= 20000LL;
      } else {
        res += y * 4LL;
        x -= 4LL;
      }
    } else {
      if (y > 20) {
        if (y > 20050) {
          res += x * 20000LL;
          y -= 20000LL;
        } else {
          res += 4LL * x;
          y -= 4LL;
        }
      }
    }
  }
  cout << res + wyn(x, y);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

Little C loves number «3» very much. He loves all things about it.

Now he is playing a game on a chessboard of size n × m. The cell in the x-th row and in the y-th column is called (x,y). Initially, The chessboard is empty. Each time, he places two chessmen on two different empty cells, the Manhattan distance between which is exactly 3. The Manhattan distance between two cells (x_i,y_i) and (x_j,y_j) is defined as |x_i-x_j|+|y_i-y_j|.

He want to place as many chessmen as possible on the chessboard. Please help him find the maximum number of chessmen he can place.

Input

A single line contains two integers n and m (1 ≤ n,m ≤ 10^9) — the number of rows and the number of columns of the chessboard.

Output

Print one integer — the maximum number of chessmen Little C can place.

Examples

Input

2 2


Output

0

Input

3 3


Output

8

Note

In the first example, the Manhattan distance between any two cells is smaller than 3, so the answer is 0.

In the second example, a possible solution is (1,1)(3,2), (1,2)(3,3), (2,1)(1,3), (3,1)(2,3).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class TH>
void _dbg(const char* sdbg, TH h) {
  cerr << sdbg << " = " << h << endl;
}
template <class TH, class... TA>
void _dbg(const char* sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << " = " << h << ", ";
  _dbg(sdbg + 1, a...);
}
const int maxn = (1e6) + 7;
const int maxk = 20;
const int inf = (1e9) + 7;
const long long LLinf = ((long long)1e18) + 7LL;
const long double eps = 1e-9;
const long long mod = 1e9 + 7;
struct Matching {
  int n, wynik;
  vector<int> odl, str, para, odw;
  const vector<vector<int> >& V;
  Matching(int N, const vector<vector<int> >& M, const vector<int>& s)
      : n(N), odl(n + 1), str(s), para(n + 1), odw(n + 1), V(M) {}
  bool DFS(int v, int ojciec, int zostalo) {
    if (zostalo < 0) return false;
    odw[v] = 1;
    if (!str[v]) {
      if (!para[v]) return (para[v] = ojciec);
      if (DFS(para[v], v, zostalo - 1)) return (para[v] = ojciec);
      return false;
    }
    for (int i = 0; i < ((int)(V[v]).size()); i++)
      if (odl[V[v][i]] == odl[v] + 1 && !odw[V[v][i]] &&
          DFS(V[v][i], v, zostalo - 1))
        return (para[v] = V[v][i]);
    return false;
  }
  int sciezki() {
    memset(&odl[0], 0, 4 * (n + 1));
    memset(&odw[0], 0, 4 * (n + 1));
    int minodl = 2 * n + 5, ilezmian = 0;
    queue<pair<int, int> > Q;
    for (int i = 1; i <= n; i++)
      if (str[i] && !para[i]) {
        Q.push(make_pair(i, 1));
        odl[i] = 1;
      }
    while (!Q.empty()) {
      int akt = Q.front().first, od = Q.front().second;
      Q.pop();
      if (!str[akt]) {
        if (para[akt]) {
          Q.push(make_pair(para[akt], od + 1));
          odl[para[akt]] = od + 1;
        } else
          minodl = min(minodl, od);
        continue;
      }
      for (int i = 0; i < ((int)(V[akt]).size()); i++)
        if (odl[V[akt][i]] == 0) {
          Q.push(make_pair(V[akt][i], od + 1));
          odl[V[akt][i]] = od + 1;
        }
    }
    for (int i = 1; i <= n; i++)
      if (!odw[i] && str[i] && !para[i]) ilezmian += DFS(i, -1, minodl - 1);
    return ilezmian;
  }
  int matching() {
    int ile = 0, akt;
    while ((akt = sciezki())) ile += akt;
    return wynik = ile;
  }
  vector<int> mpn() {
    vector<int> stan(n + 1, 0);
    queue<int> Q;
    for (int i = 1; i <= n; i++) {
      if (!para[i]) {
        stan[i] = 1;
        Q.push(i);
      }
    }
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      for (auto el : V[x]) {
        if (stan[el] == 0) {
          stan[el] = -1;
          stan[para[el]] = 1;
          Q.push(para[el]);
        }
      }
    }
    vector<int> odp;
    for (int i = 1; i <= n; i++)
      if (stan[i] == 1 || (!stan[i] && str[i])) odp.push_back(i);
    return odp;
  }
};
vector<pair<int, int> > mov = {{-1, -2}, {-1, 2}, {-2, -1}, {-2, 1},
                               {1, -2},  {1, 2},  {2, -1},  {2, 1},
                               {3, 0},   {0, 3},  {-3, 0},  {0, -3}};
int A, B;
int numer(int a, int b) { return (a - 1) * B + b; }
long long wyn(int a, int b) {
  A = a;
  B = b;
  vector<vector<int> > V;
  vector<int> cnt;
  V.resize(numer(a, b) + 10);
  cnt.resize(numer(a, b) + 10);
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++) {
      for (auto s : mov) {
        int ii = i + s.first;
        int jj = j + s.second;
        if (1 <= ii && ii <= a && 1 <= jj && jj <= b) {
          V[numer(i, j)].push_back(numer(ii, jj));
        }
      }
      cnt[numer(i, j)] = (i + j) & 1;
    }
  Matching xd(numer(a, b) + 5, V, cnt);
  return xd.matching() * 2LL;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long x, y;
  cin >> x >> y;
  long long res = 0LL;
  if (x > y) swap(x, y);
  if (x == 1LL) {
    long long cnt = y / 6;
    y -= cnt * 6;
    cout << cnt * 6 + max(0LL, y - 3) * 2LL;
    return 0;
  }
  while (x * y > 1000) {
    if (x > 20) {
      if (x > 20050) {
        res += 20000LL * y;
        x -= 20000LL;
      } else {
        res += y * 4LL;
        x -= 4LL;
      }
    } else {
      if (y > 20) {
        if (y > 20050) {
          res += x * 20000LL;
          y -= 20000LL;
        } else {
          res += 4LL * x;
          y -= 4LL;
        }
      }
    }
  }
  cout << res + wyn(x, y);
  return 0;
}

```