Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.

The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.

Input

The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).

Output

Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range. 

Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.

Examples

Input

5 3 3 1 1


Output

3.7677669529663684 3.7677669529663684 3.914213562373095


Input

10 5 5 5 15


Output

5.0 5.0 10.0
'''input
1 2 3
4 5
6 7 8
45 9
'''

import sys
pow2 = pow              # for modular expo pow2(base, n, mod)
from math import *
from time import time
from collections import defaultdict
from bisect import bisect_right, bisect_left
from string import ascii_lowercase as lcs
from string import ascii_uppercase as ucs
from fractions import Fraction, gcd
from decimal import Decimal, getcontext
from itertools import product, permutations, combinations
#getcontext().prec = 500
#sys.setrecursionlimit(30000)

# What doesn't challenge you, makes you weaker.

mod = 10**9 + 7
INF = sys.maxint

raw_input = lambda: sys.stdin.readline().rstrip()
die = lambda : exit(0)
flush= lambda : sys.stdout.flush()

r_s = lambda: raw_input().strip()                   #read str
r_ss = lambda: raw_input().split()          #read stringss
r_i = lambda: int(raw_input())              #read int
r_is = lambda: map(int, raw_input().split())#read ints
r_f = lambda: float(raw_input())            #read float
r_fs = lambda: map(Decimal, raw_input().split()) #read floats

# For effieciently taking lots of queries with each query in a separate line
# having integer values separated by space. A 2d list is made with each row
# corresponding to line. Each row is a list with all the values.
# Use this for fast I/O by taking all input in one go.
r_qs = lambda: map(lambda s:map(int, s.split(' ')), sys.stdin.readlines())

'''-------------------------------------------------------------------'''

# Call it as dbg(varName=varName)
def dbg(**kwargs):
    print "Value of '" + kwargs.keys()[0] + "' is", kwargs[kwargs.keys()[0]]

import math
def main():
    r, ax,ay, bx, by = r_is()
    dist  = (ax-bx)**2 + (ay-by)**2
    if dist>=r**2:
        print "%.13f %.13f %.13f" %(ax, ay, r)
        return
    if ax==bx and ay==by:
        dx = ax+r
        dy = ay
    else:
        cx = ax + r*((bx-ax)/sqrt((bx-ax)**2+(by-ay)**2))
        cy = ay + r*((by-ay)/sqrt((bx-ax)**2+(by-ay)**2))
        dx = 2*ax-cx
        dy = 2*ay-cy
    rad = sqrt((dx-bx)**2 + (dy-by)**2)/2.0
    print "%.13f %.13f %.13f" %((dx+bx)/2.0, (dy+by)/2.0, rad)

if __name__ == '__main__':
    local = True
    # local = False
    if local:
        try:
            sys.stdin = open('input.txt')
            print 'Running at local.'
            for i in range(r_i()):
                main()
        except IOError:
            main()
    else:
        main()


