Little C loves number «3» very much. He loves all things about it.

Now he has a positive integer n. He wants to split n into 3 positive integers a,b,c, such that a+b+c=n and none of the 3 integers is a multiple of 3. Help him to find a solution.

Input

A single line containing one integer n (3 ≤ n ≤ 10^9) — the integer Little C has.

Output

Print 3 positive integers a,b,c in a single line, such that a+b+c=n and none of them is a multiple of 3.

It can be proved that there is at least one solution. If there are multiple solutions, print any of them.

Examples

Input

3


Output

1 1 1

Input

233

Output

77 77 79
#import resource
#import sys
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**26)
#sys.setrecursionlimit(0x1000000)
from sys import stdin, stdout
mod=(10**9)+7
mod1=mod-1
def modinv(n,p):
    return pow(n,p-2,p)
def ncr(n,r,p):
    t=((fact[n])*(modinv(fact[r],p)%p)*(modinv(fact[n-r],p)%p))%p
    return t
def GCD(x, y):
   while(y):
       x, y = y, x % y
   return x
def BS(arr, l, r, x):
    if r >= l:
        mid = l + (r - l)/2
        if arr[mid] == x:
            return mid
        elif arr[mid] > x:
            return BS(arr, l, mid-1, x)
        else:
            return BS(arr, mid+1, r, x)
    else:n -1
from bisect import bisect_left as bl
from bisect import bisect_right as br
import itertools
import math
import heapq
from random import randint as rn
from Queue import Queue as Q
def comp(x,y):
    if(x[0]<y[0]):
        return -1
    elif(x[0]==y[0]):
        if(x[1]<y[1]):
            return -1
        else:
            return 1
    else:
        return 1
import heapq
#credits:TDS115/yolaaa kyonki plagarism me pakda gaya randi
"""*********************************************************************************"""
a=input()
w=a-1
if((w-1)%3!=0):
    print 1,1,a-2
else:
    print 1,2,a-3
