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
def Compute(n):
    k = (n - 4) / 9
    if checkForInteger(k):
        a = 3 * k + 1
        b = 3 * k + 2
        c = a
        print(int(a), int(b), int(c))
        return
    k = (n - 5) / 9
    if checkForInteger(k):
        a = 3 * k + 1
        b = 3 * k + 2
        c = b
        print(int(a), int(b), int(c))
        return
    k = (n - 3) / 9
    if checkForInteger(k):
        a = n / 3
        b = a
        c = b
        print(int(a), int(b), int(c))
        return
    k = (n - 6) / 9
    if checkForInteger(k):
        a = n / 3
        b = a
        c = b
        print(int(a), int(b), int(c))
        return
    k = (n - 6) // 9
# Case 1
    a = (3 * k) + 1
    b = (n - a) / 2
    if checkForInteger(b) and b % 3 != 0:
        print(int(a), int(b), int(b))
        return
    c = n - (2 * a)
    if c % 3 != 0:
        print(int(a), int(a), int(c))
        return
# Case 2
    a = (3 * k) + 2
    b = (n - a) / 2
    if checkForInteger(b) and b % 3 != 0:
        print(int(a), int(b), int(b))
        return
    c = n - (2 * a)
    if c % 3 != 0:
        print(int(a), int(a), int(c))
        return

def checkForInteger(x):
    stri = str(x)
    index = stri.index('.')
    test = stri[index + 1:]
    zero = 0
    for i in test:
        if i == '0':
            zero += 1
    if zero == len(test):
        return True
    else:
        return False

n = int(input())
Compute(n)