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.
t = int(input())
for _ in range(t):
    a, b = map(int, input().split())
    if 2*a-b<0:
        print(a)
    elif 2*b-a<0:
        print(b)
    else:
        x, y = (2*a-b), (2*b-a)
        if x % 3 == 0 and y % 3 == 0:
            print((2*a-b)//3+(2*b-a)//3)
        else:
            lu = 2*(x//3)+y//3+1<=a and (x//3)+2*(y//3+1)<=b
            ru = 2*(x//3+1)+y//3+1<=a and (x//3+1)+2*(y//3+1)<=b
            ld = 2*(x//3)+y//3<=a and (x//3)+2*(y//3)<=b
            rd = 2*(x//3+1)+y//3<=a and (x//3+1)+2*(y//3)<=b
            if ru:
                print((x//3+1)+(y//3+1))
            elif lu or ru:
                print((x//3)+(y//3+1))
            else:
                print((x//3)+(y//3))