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.
import java.util.*;

public class A{

    static boolean check(int a, int b, int mid){
        if(a-2*mid<b-mid)
            return true;
        return false;
    }

    static int binarySearch(int l, int u, int a, int b){
        if(l>u)
            return -1;

        int mid = (int)((long)l+u)/2;

        boolean check1, check2;

        check1 = check(a,b,mid);
        check2 = check(a,b,mid-1);
        if(check1 && !check2)
            return mid;
        else if(check1)
            return binarySearch(l, mid-1, a, b);
        else
            return binarySearch(mid+1, u, a, b);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();

        for(int t=0; t<T; ++t){
            int a = sc.nextInt();
            int b = sc.nextInt();

            if(b>a){
                int temp = a;
                a = b;
                b = temp;
            }
            int count = 0;
            while(b>0 && a>1){
                int mid = binarySearch(1, a/2 +1, a, b);
                if(mid<0)
                    break;
                a -= 2*mid;
                b -= mid;

                if(a>=0 && b>=0)
                    count += mid;
                else
                    count += mid-1;

                int temp = a;
                a = b;
                b = temp;

                // System.out.println("A: "+a+" B: "+b);
            }
            System.out.println(count);
        }
        sc.close();
    }
}