human
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....
Using python can you solve the prior task?
assistant
#Here is Python program to print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.

from __future__ import division, print_function
import bisect
import math
import heapq
import itertools
import sys
from collections import deque
from atexit import register
from collections import Counter
from functools import reduce
sys.setrecursionlimit(100000)
if sys.version_info[0] < 3:
    from io import BytesIO as stream
else:
    from io import StringIO as stream
 
 
if sys.version_info[0] < 3:
    class dict(dict):
        """dict() -> new empty dictionary"""
        def items(self):
            """D.items() -> a set-like object providing a view on D's items"""
            return dict.iteritems(self)
 
        def keys(self):
            """D.keys() -> a set-like object providing a view on D's keys"""
            return dict.iterkeys(self)
 
        def values(self):
            """D.values() -> an object providing a view on D's values"""
            return dict.itervalues(self)
 
    input = raw_input
    range = xrange
 
    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip
 
 
def sync_with_stdio(sync=True):
    """Set whether the standard Python streams are allowed to buffer their I/O.
 
    Args:
        sync (bool, optional): The new synchronization setting.
 
    """
    global input, flush
 
    if sync:
        flush = sys.stdout.flush
    else:
        sys.stdin = stream(sys.stdin.read())
        input = lambda: sys.stdin.readline().rstrip('\r\n')
 
        sys.stdout = stream()
        register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))

def dd():
    return map(int,input().split())    
def arr():
    return list(map(int,input().split()))
def twinSort(X,Y):
    #sorting X wrt Y
    return [x for _,x in sorted(zip(Y,X))]

def solve():
    a,b=dd()

    print(a-1,b)

    



def main():
    testCase=1
    if testCase:
        for _ in range(int(input())):
            solve()
    else:
        solve()
    
if __name__ == '__main__':
    sync_with_stdio(False)
    main()