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.
# May the Speedforce be with us...
'''
problem statement achhe se padhna hai
age se bhi dekhna hai, piche se bhi dekhna hai
'''
#from math import gcd,ceil,sin,radians,cos
#from bisect import insort
#from collections import defaultdict as dd
def lcm(a,b):
	return (a*b)//gcd(a,b)
def inin():
	return int(input())
def inar():
	return list(map(int,input().split()))
def ar(element,size):
	return [element for i in range(size)]
def yes():
	print('Yes')
def no():
	print('No')
'''
def countDivisibles(A, B, M): #count numbers in the range[A,B] which are divisible by M
    if (A % M == 0): 
        return ((B // M) - (A // M)) + 1
    return ((B // M) - (A // M)) 
	
def  countSetBits(n): #count the number of set bits in the binary representation of an integer
    count = 0
    while (n): 
        count += n & 1
        n >>= 1
    return count
def digitsum(num): #return the sum of dogots of a number
	su=0
	while(num):
		su+=num%10
		num//=10
	return su
'''


Testcase=inin()
for _ in range(Testcase):
	a,b=inar()
	print(min(a,b,(a+b)//3))

	
