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.
import sys
from collections import defaultdict as dd
from collections import deque
from fractions import Fraction as f
def eprint(*args):
    print(*args, file=sys.stderr)
zz=1
from math import *
import copy
#sys.setrecursionlimit(10**6)
if zz:
	input=sys.stdin.readline
else:	
	sys.stdin=open('input.txt', 'r')
	sys.stdout=open('all.txt','w')
def li():
	return [int(x) for x in input().split()]
def fi():
	return int(input())
def si():
	return list(input().rstrip())	
def mi():
	return 	map(int,input().split())	
def gh():
	sys.stdout.flush()
def bo(i):
	return ord(i)-ord('a')
from copy import *
from bisect import *	

from itertools import permutations 


def cal(l):
	return l//4+(l-l//2)

t=int(input())
while t>0:
	t-=1
	a,b=mi()
	if a<b:
		a,b=b,a
	
	if 2*b<=a:
		print(b)
	else:
		print((a+b)//3)		


