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 <stdio.h>
#include <algorithm>
#include <iostream>
#include <limits.h>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#include <cstdio>
#include <list>
#define ll long long
#define rep2(i,a,b) for(ll i=a;i<=b;i++)
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep3(i,a,b) for(ll i=a;i>=b;i--)
#define REP(e,v) for(auto e:v)
#define pii pair<int,int>
#define pll pair<ll,ll>
#define tii tuple<int,int,int>
#define pq priority_queue<int>
#define pqg priority_queue<int,vector<int>,greater<int>>
#define pb push_back
#define edge(v,a,b) v[a].pb(b);v[b].pb(a);
#define vec vector<int>
#define vecll vector<ll>
#define vecstr vector<string>
#define vecch vector<char>
#define vecpii vector<pii>
#define bs binary_search
#define All(c) (c).begin(),(c).end()
using namespace std;
int in() {int x;scanf("%d",&x);return x;}
string strin(){string s;cin>>s;return s;}
char chin(){char c;cin>>c;return c;}
ll lin() {ll x;scanf("%lld",&x);return x;}

int main(){
    ll n=lin(),m=lin();
    if(n>m*2){
        cout<<m<<endl;
        return 0;
    }
    cout<<(m-2*n)/4+n<<endl;
}