Polycarp plays a well-known computer game (we won't mention its name). In this game, he can craft tools of two types — shovels and swords. To craft a shovel, Polycarp spends two sticks and one diamond; to craft a sword, Polycarp spends two diamonds and one stick.

Each tool can be sold for exactly one emerald. How many emeralds can Polycarp earn, if he has a sticks and b diamonds?

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

The only line of each test case contains two integers a and b (0 ≤ a, b ≤ 10^9) — the number of sticks and the number of diamonds, respectively.

Output

For each test case print one integer — the maximum number of emeralds Polycarp can earn.

Example

Input


4
4 4
1000000000 0
7 15
8 7


Output


2
0
7
5

Note

In the first test case Polycarp can earn two emeralds as follows: craft one sword and one shovel.

In the second test case Polycarp does not have any diamonds, so he cannot craft anything.
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long lli;
typedef double lld;
#define mod 1000000007
#define mod2 998244353
#define fori(n) for (int i = 0; i < n; i++)
#define forj(n) for (int j = 0; j < n; j++)
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define fast ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define endl "\n"
#define sortvec(a) sort(a.begin(), a.end())
template <typename T, typename U>
ostream &operator<<(ostream &out, const pair<T, U> &p){out << "(" << p.first << ", " << p.second << ")";return out;}
template <template <typename, typename...> class ContainerType, typename ValueType, typename... Args>
void print(const ContainerType<ValueType, Args...> &c){for (const auto &v : c)cout << v << ' ';cout << '\n';}
void print(){cout<<'\n';}
template <typename T, typename... Args>
void print(T a, Args... args){cout << a << " ";print(args...);}
lli llipower(lli x, lli y, lli p = LLONG_MAX){lli res = 1;x = x % p;while (y > 0){if (y & 1)res = (res * x) % p;y = y >> 1;x = (x * x) % p;}return res;}

//8====================================
int main(){
    int t;
    cin>>t;
    while(t--){
        lli a,b;
        cin>>a>>b;
        if(b>a) swap(a,b);
        lli d=a-b;
        lli ans=0;
        ans=min({d,2*a,b});
        a-=ans;
        b-=ans;
        ans+=2*(a/3,b/3);
        cout<<ans<<endl;




    }
}