Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible. 

The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.

Input

The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.

Output

On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.

Examples

Input

3 1


Output

2 1


Input

2 4


Output

3 2

Note

In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point. 

If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points.
"""Things to do if you are stuck:-
1.Read the problem statement again, maybe you've read something wrong.
2.See the explanation for the sample input .
3.If the solution is getting too complex in cases where no. of submissions 
  are high ,then drop that idea because there is something simple which you are
  missing.
4.Check for runtime errors if unexpected o/p is seen.
5.Check on edge cases before submitting.
6.Ensure that you have read all the inputs before returning for a test case.
7.Try to think of brute force first if nothing is striking.
8.Take more examples
9.Don't give up , maybe you're just one statement away! """
#pyrival orz
import os
import sys
import math
from io import BytesIO, IOBase


input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))

############ ---- Dijkstra with path ---- ############
def dijkstra(start, distance, path, n):
    # requires n == number of vertices in graph,
    # adj == adjacency list with weight of graph

    visited = [False for _ in range(n)]  # To keep track of vertices that are visited
    distance[start] = 0  # distance of start node from itself is 0

    for i in range(n):
        v = -1  # Initialize v == vertex from which its neighboring vertices' distance will be calculated
        for j in range(n):
            # If it has not been visited and has the lowest distance from start
            if not visited[v] and (v == -1 or distance[j] < distance[v]):
                v = j

        if distance[v] == math.inf:
            break

        visited[v] = True  # Mark as visited

        for edge in adj[v]:
            destination = edge[0]  # Neighbor of the vertex
            weight = edge[1]  # Its corresponding weight

            if distance[v] + weight < distance[destination]:  # If its distance is less than the stored distance
                distance[destination] = distance[v] + weight  # Update the distance
                path[destination] = v  # Update the path
        

def gcd(a, b):
    if b == 0:
        return a
    else:
        return gcd(b, a%b)

def lcm(a, b):
    return (a*b)//gcd(a, b)

def ncr(n, r):
    return math.factorial(n)//(math.factorial(n-r)*math.factorial(r))

def npr(n, r):
    return math.factorial(n)//math.factorial(n-r)

def seive(n):
    primes = [True]*(n+1)
    ans = []

    for i in range(2, n):
        if not primes[i]:
            continue

        j = 2*i
        while j <= n:
            primes[j] = False
            j += i

    for p in range(2, n+1):
        if primes[p]:
            ans += [p]

    return ans

def factors(n):
    factors = []

    x = 1
    while x*x <= n:

        if n % x == 0:
            if n // x == x:
                factors.append(x)
            else:
                factors.append(x)
                factors.append(n//x)

        x += 1

    return factors

# Functions: list of factors, seive of primes, gcd of two numbers,
# lcm of two numbers, npr, ncr

def main():
    try:
        n, m = invr()
        print(max(n, m) - 1, min(n, m))

    except Exception as e:
        print(e)

        

# region fastio

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

if __name__ == "__main__":
    main()