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 solve(st, di, em):
    if (st <= 0 or di < 2) and (di <= 0 or st < 2):
        return em

    if (st >= 1 and di >= 2) and (di >= 1 and st >= 2):
        if st > di:
            return solve(st - 2, di - 1, em + 1)
        else:
            return solve(st - 1, di - 2, em + 1)
    elif (st > 0 and di >= 2):
        return solve(st - 1, di - 2, em + 1)
    elif (di > 0 and st >= 2):
        return solve(st - 2, di - 1, em + 1)
    else:
        pass


# def solve2(a, b):
#     if min(a, b) * 2 <= max(a, b):
#         return min(a, b)
#     mi = min(a, b)
#     other_max = 2 * mi
#     sub = int((other_max - max(a, b)) / 2)
#     dif = abs(max(a, b) - mi)
#     return mi - sub + dif


for case in range(t):
    s, d = map(int, input().split())

    ma = max(s, d)
    mi = min(s, d)
    dif = ma - mi
    if dif > mi:
        print(mi)
    else:
        ems = ma - mi
        same = mi - dif
        ems += int(same * 2 / 3)
        print(ems)

