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.
from math import *
from copy import *
from string import *				# alpha = ascii_lowercase
from random import *
from sys import stdin,stdout
from sys import maxsize
from operator import *				# d = sorted(d.items(), key=itemgetter(1))
from itertools import *
from collections import Counter		# d = dict(Counter(l))
import math
import math
import time
from queue import Queue
# PradeepGhosh_2017076

def seive(n):
	l=[True]
	l=l*(n+1)
	for i in range(2,int(sqrt(n))+1):
		if(l[i]==True):
			val=i*i
			while(val<len(l)):
				l[val]=False
				val+=i
	prime=[]
	for i in range(2,len(l)):
		if(l[i]==True):
			prime.append(i)
	return prime

def dp(l,i,n,val,ans):
	if(i>=n or val<0):
		return 1000000
	elif(val==0):
		return 0
	else:
		if(ans[i]==1000000):
			ans[i]=min(min(1+dp(l,i+1,n,val-l[i],ans),1+dp(l,i,n,val-l[i],ans)),dp(l,i+1,n,val,ans))
		return ans[i]
from queue import Queue
if __name__ == '__main__':
	t=int(input())
	for i in range(t):
		a,b=map(int,input().split(" "))
		if(a>=b*2):
			print(b)
		elif(b>=a*2):
			print(a)
		else:
			print((a+b)//3)
	



