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.
for _ in range(int(input())):
    a, b = map(int, input().split())
    counter = 0
    # print(a, b)
    while a > 0 and b > 0:
        # print('!', a, b)
        if a >= 2 * b:
            # print('!')
            counter = b
            break
        elif b >= 2 * a:
            # print('!')
            counter = a
            break
        elif a == b:
            # print('!')
            if a % 3 == 0:
                counter += (a // 3) * 2
            if a % 3 == 1:
                # print('!')
                counter += (a // 3) * 2
            if a % 3 == 2:
                counter += (a // 3) * 2 + 1
            break
        elif a > b:
            nexus = a - b
            if nexus * 2 <= a:
                counter += nexus
                a -= nexus * 2
                b -= nexus
            else:
                counter += a // 2
                break
        elif a < b:
            nexus = b - a
            if nexus * 2 <= b:
                counter += nexus
                b -= nexus * 2
                a -= nexus
            else:
                counter += b // 2
                break
        if a == 1 and b == 1:
            break
    print(counter)

