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.
/**
 * @author egaeus
 * @mail sebegaeusprogram@gmail.com
 * @veredict Not sended
 * @url <https://codeforces.com/problemset/problem/CF1366A>
 * @category ?
 * @date 11/06/2020
 **/

import java.io.*;
import java.util.*;

import static java.lang.Integer.*;
import static java.lang.Math.*;

public class CF1366A {
    public static void main(String args[]) throws Throwable {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int T = parseInt(in.readLine());
        for (int t = 0; t < T; t++) {
            StringTokenizer st = new StringTokenizer(in.readLine());
            int a = parseInt(st.nextToken()), b = parseInt(st.nextToken());
            /*int s = Math.min(a / 2, b), s2 = Math.min(b / 2, a);
            int A = a - s * 2;
            int B = b - s;
            int solutionA = s + (Math.min(B / 2, A));
            A = a - s2;
            B = b - s2 * 2;
            int solutionB = s2 + (Math.min(A / 2, B));
            System.out.println(Math.max(solutionA, solutionB));*/
            System.out.println(Math.min(Math.min(a, b), (a+b)/3));
        }
    }
}
