Alice and Bob play ping-pong with simplified rules.

During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.

The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.

Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.

Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.

Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.

Calculate the resulting number of Alice's and Bob's wins.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.

Output

For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.

Example

Input


3
1 1
2 1
1 7


Output


0 1
1 1
0 7

Note

In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.

In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.

In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
import time,math as mt,bisect as bs,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
    return int(stdin.readline())
def IP(): # to take tuple as input
    return map(int,stdin.readline().split())
def L(): # to take list as input
    return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
    return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
    return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
    return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
    if a==0:
        return b
    elif b==0:
        return a
    if a>b:
        return gcd(a%b,b)
    else:
        return gcd(a,b%a)
def bfs(adj,v): # a schema of bfs
    visited=[False]*(v+1)
    q=deque()
    while q:
        pass
def setBit(n):
    count=0
    while n!=0:
        n=n&(n-1)
        count+=1
    return count
mx=10**7
spf=[mx]*(mx+1)
def readTree(n,e): # to read tree
    adj=[set() for i in range(n+1)]
    for i in range(e):
        u1,u2=IP()
        adj[u1].add(u2)
    return adj
def sieve():
    li=[True]*(10**3+5)
    li[0],li[1]=False,False
    for i in range(2,len(li),1):
        if li[i]==True:
            for j in range(i*i,len(li),i):
                li[j]=False
    prime,cur=[0]*200,0
    for i in range(10**3+5):
        if li[i]==True:
            prime[cur]=i
            cur+=1
    return prime
def SPF():
    mx=(10**6+1)
    spf[1]=1
    for i in range(2,mx):
        if spf[i]==1e9:
            spf[i]=i
            for j in range(i*i,mx,i):
                if i<spf[j]:
                    spf[j]=i
    return
def prime(n,d):
    prm=set()
    while n!=1:
        prm.add(spf[n])
        n=n//spf[n]
    for ele in prm:
        d[ele]=d.get(ele,0)+1
    return
#####################################################################################
mod=998244353
def solve():
    x,y=IP()
    print(x-1,y)

    return

t=II()
for i in range(t):
    solve()

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