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())
def search(num1,num2,lst):

    for i in lst:
        if(i>num1 or i<num2):
            return 1
    return 0

for _ in range(t):
    a,b=map(int,input().split())
    if(a==0 or b==0):
        print(0)
    else:
        if(a>b):
            if(a>=2*b):
                print(b)
            else:
                ct=0
                ct+=a-b
                a1=2*b-a
                b1=2*b-a
                if(a1%3==0 or a1%3==1):
                    ct+=(a1//3)*2
                elif(a1%3==2):
                    ct+=(((a1//3)*2)+1)
                print(ct)
        else:
            a1,b1=b,a
            a,b=a1,b1
            if (a >= 2 * b):
                print(b)
            else:
                ct = 0
                ct += a - b
                a1 = 2 * b - a
                b1 = 2 * b - a
                if (a1 % 3 == 0 or a1 % 3 == 1):
                    ct += (a1 // 3) * 2
                elif (a1 % 3 == 2):
                    ct += (((a1 // 3) * 2) + 1)
                print(ct)









