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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include "debug.h"
#else
#define dump(...)
#endif
#define endl "\n"
#define ll long long
#define FOR(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define REP(i, x) for (int i = 0; i < (int)(x); i++)
#define REPS(i, x) for (int i = 1; i <= (int)(x); i++)
#define RREP(i, x) for (int i = ((int)(x)-1); i >= 0; i--)
#define RREPS(i, x) for (int i = ((int)(x)); i > 0; i--)
#define INF 2147483647
#define LLINF 9223372036854775807LL
#define vi vector<int>
#define vvi vector<vector<int>>
#define pi pair<int, int>
#define ALL(a) (a).begin(), (a).end()
#define BIT(n) (1LL << (n))
#define TOUPPER(s) transform(s.begin(), s.end(), s.begin(), ::toupper)
#define TOLOWER(s) transform(s.begin(), s.end(), s.begin(), ::tolower)
#define FILLZERO(s, n)                      \
    do {                                    \
        ostringstream os;                   \
        os << setw(n) << setfill('0') << s; \
        s = os.str();                       \
    } while (0);
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end()) //sortしてからつかうこと
#define INIT                               \
    std::cin.tie(nullptr);                 \
    std::ios_base::sync_with_stdio(false); \
    cout << std::fixed << std::setprecision(10);

constexpr ll MOD = 1e9 + 7;
int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
ll A, B, C, D, E, F, G, H, N, M, L, K, P, Q, R, W, X, Y, Z;
string S, T;
ll ans = 0;

template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
    for (T &x : vec)
        is >> x;
    return is;
}
template <class T>
inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return 1;
    }
    return 0;
}

signed main() {
    INIT;
    cin >> N >> M;
    M /= 2;
    if (N > (N + M) / 2LL)
        cout << M << endl;
    else
        cout << (N + M) / 2LL << endl;
}