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.
#!/usr/bin/python3

import getopt
import sys
import math

if __name__ == "__main__":
    onlyCase = -1
    verbose = False
    
    try:
         opts, args = getopt.getopt(sys.argv[1:], "vc:",
                                   ["case="])
    except getopt.GetoptError as err:
        print (str(err)) 
        sys.exit(2)

    for o, a in opts:
        if o in ("-c", "--case"):
            onlyCase = int(a)
        elif o  == "-v":
            verbose = True
        else: sys.exit()

    
    ncases = int(input())
    for c in range(1, ncases + 1):
        A, B = [int(x) for x in input().split()]

        if onlyCase != -1 and onlyCase != c:
            continue

        if verbose:
            print("A= %d B=%d" % (A, B))
        
        if A < B:
            A, B = B, A

        x1 = min ((2*A - B) // 3, B)

        A = A - (2*x1)
        B = B - x1

        x2 = min(A, B//2)

        if verbose:
            print("x1= %d x2=%d" % (x1, x2))
        
        print("%d" % (x1+x2))





        

        




