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.
//package er89;

import java.util.Scanner;

public class first {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc =new Scanner(System.in);
		int t = sc.nextInt();
		while(t-->0)
		{
			int a = sc.nextInt();
			int b = sc.nextInt();
			if(a==0 || b==0)
			{
				System.out.println(0);
				continue;
			}
			int sho = Math.min(a/2, b);
			int swo = Math.min(b/2, a);
			int al1 = a-(sho*2);
			int al2 =  (a)-swo;
			int bl1 = (b) - sho;
			int bl2 = (b)-(swo*2);
			if(bl1>2)
			{
				sho--;
				bl1+=1;
				int add = Math.min(bl1/2, al1+2);
				bl1-=(add*2);
				al1-=(add);
				sho+=add;
				
			}
			else if(bl1==2 && al1>=1)
			{
				sho+=1;
			}
			if(al2>2)
			{
				swo--;
				al2+=1;
				int add = Math.min(al2/2,bl2+2);
				al2-=(add*2);
				bl2-=(add);
				swo+=add;
			}
			else if(al2==2 && bl2>=1)
			{
				swo+=1;
			}
			
			System.out.println(Math.max(sho, swo));
			
		}

	}

}
