Problem:
Score : 300 points

Problem Statement

Snuke loves puzzles.

Today, he is working on a puzzle using S- and c-shaped pieces.
In this puzzle, you can combine two c-shaped pieces into one S-shaped piece, as shown in the figure below:

Snuke decided to create as many Scc groups as possible by putting together one S-shaped piece and two c-shaped pieces.

Find the maximum number of Scc groups that can be created when Snuke has N S-shaped pieces and M c-shaped pieces.

Constraints

1 ≤ N,M ≤ 10^{12}

Input

The input is given from Standard Input in the following format:

N M

Output

Print the answer.

Sample Input 1

1 6

Sample Output 1

2

Two Scc groups can be created as follows:

Combine two c-shaped pieces into one S-shaped piece

Create two Scc groups, each from one S-shaped piece and two c-shaped pieces

Sample Input 2

12345 678901

Sample Output 2

175897

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define all(x) (x).begin(),(x).end()
typedef long long ll;
#define rep(i,n) for(ll i=0, i##_len=(n); i<i##_len; ++i)
#define REP(i,num,n) for(ll i=num, i##_len=(n); i<i##_len; ++i)
#define repprev(i,a,b) for(ll i=b-1;i>=a;i--)
#define reprev(i,n) repprev(i,0,n)
using namespace std;
#define sz(x) ((int)(x).size())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define MEMSET(v, h) memset((v), h, sizeof(v))
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<class T> int former(const vector<T> &v, T x){
   return upper_bound(v.begin(),v.end(),x) - v.begin() - 1;
}
template<class T> int latter(const vector<T> &v, T x){
    return lower_bound(v.begin(),v.end(),x) - v.begin();
}
#define pb push_back
#define mp make_pair
#define y0 y3487465
#define y1 y8687969
#define j0 j1347829
#define j1 j234892
#define BIT_FLAG_0 (1<<0) // 0000 0000 0000 0001
#define BIT_FLAG_1 (1<<1) // 0000 0000 0000 0010
#define BIT_FLAG_2 (1<<2) // 0000 0000 0000 0100
#define BIT_FLAG_3 (1<<3) // 0000 0000 0000 1000
#define BIT_FLAG_4 (1<<4) // 0000 0000 0001 0000
#define BIT_FLAG_5 (1<<5) // 0000 0000 0010 0000
#define BIT_FLAG_6 (1<<6) // 0000 0000 0100 0000
#define BIT_FLAG_7 (1<<7) // 0000 0000 1000 0000
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
const ll LLINF = 1LL<<60;
const int INTINF = 1<<30;

const int MAX = 510000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

struct UnionFind {
    vector<ll> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(ll n) : par(n, -1) { }
    void init(ll n) { par.assign(n, -1); }

    ll root(ll x) {
        if (par[x] < 0) return x;
        else return par[x] = root(par[x]);
    }

    bool issame(ll x, ll y) {
        return root(x) == root(y);
    }

    bool merge(ll x, ll y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (par[x] > par[y]) swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }

    ll size(ll x) {
        return -par[root(x)];
    }
};

template <typename T>
vector<T> dijkstra(int s,vector<vector<pair<int, T> > > & G){
   const T INF = numeric_limits<T>::max();
   using P = pair<T, int>;
   int n=G.size();
   vector<T> d(n,INF);
   vector<int> b(n,-1);
   priority_queue<P,vector<P>,greater<P> > q;
   d[s]=0;
   q.emplace(d[s],s);
   while(!q.empty()){
     P p=q.top();q.pop();
     int v=p.second;
     if(d[v]<p.first) continue;
     for(auto& e:G[v]){
       int u=e.first;
       T c=e.second;
       if(d[u]>d[v]+c){
         d[u]=d[v]+c;
         b[u]=v;
         q.emplace(d[u],u);
       }
     }
   }
   return d;
}

vector<vector<int> > bfs(vector<string> &s,int sy,int sx,char wall,int dir){
   int h=s.size(),w=s.front().size();
   vector<vector<int> > dp(h,vector<int>(w,-1));
   using P = pair<int, int>;
   queue<P> q;

   dp[sy][sx]=0;
   q.emplace(sy,sx);

   int dy[]={1,-1,0,0,1,1,-1,-1};
   int dx[]={0,0,1,-1,1,-1,1,-1};
   auto in=[&](int y,int x){return 0<=y&&y<h&&0<=x&&x<w;};

   while(!q.empty()){
     int y,x;
     tie(y,x)=q.front();q.pop();
     for(int k=0;k<dir;k++){
       int ny=y+dy[k],nx=x+dx[k];
       if(!in(ny,nx)||s[ny][nx]==wall) continue;
       if(~dp[ny][nx]) continue;
       dp[ny][nx]=dp[y][x]+1;
       q.emplace(ny,nx);
     }
   }

   return dp;
}

int64_t power(int64_t x, int64_t n, int64_t mod) {
   int64_t ret = 1;
   while(n > 0) {
     if(n & 1) (ret *= x) %= mod;
     (x *= x) %= mod;
     n >>= 1;
   }
   return ret;
}

vector<int> sieve_of_eratosthenes(int n) {
   vector<int> primes(n);
   for (int i = 2; i < n; ++i)
      primes[i] = i;
    for (int i = 2; i*i < n; ++i)
      if (primes[i])
        for (int j = i*i; j < n; j+=i)
          primes[j] = 0;
    return primes;
}

std::vector<ll> divisor(ll n)//nの約数を列挙
{
 std::vector<ll> ret;
 for(ll i=1 ; i*i<=n ; ++i)
  {
   if(n%i == 0)
    {
     ret.push_back(i);
     if(i!=1 && i*i!=n)
     {
        ret.push_back(n/i);
      }
    }
  }
  return ret;
}

const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};


int main(void) {
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll N,M;
  cin >> N >> M;
  ll scc = min(N, M / 2);
  ll cccc= (M - 2 * scc);
  cout << scc + cccc << endl;
}