Problem: Tom loves vowels, and he likes long words with many vowels. His favorite words are vowelly words. We say a word of length k is vowelly if there are positive integers n and m such that n⋅ m = k and when the word is written by using n rows and m columns (the first row is filled first, then the second and so on, with each row filled from left to right), every vowel of the English alphabet appears at least once in every row and every column.

You are given an integer k and you must either print a vowelly word of length k or print -1 if no such word exists.

In this problem the vowels of the English alphabet are 'a', 'e', 'i', 'o' ,'u'.

Input

Input consists of a single line containing the integer k (1≤ k ≤ 10^4) — the required length.

Output

The output must consist of a single line, consisting of a vowelly word of length k consisting of lowercase English letters if it exists or -1 if it does not.

If there are multiple possible words, you may output any of them.

Examples

Input


7


Output


-1


Input


36


Output


agoeuioaeiruuimaeoieauoweouoiaouimae

Note

In the second example, the word "agoeuioaeiruuimaeoieauoweouoiaouimae" can be arranged into the following 6 × 6 grid:

<image>

It is easy to verify that every row and every column contain all the vowels.

Solution (in PYTHON3):

Solution:
# Lets goto the next level 
# AIM Specialist at CF *__* asap 
# template taken from chaudhary_19  
# Remember you were also a novice when you started, 
# hence never be rude to anyone who wants to learn something
# Never open a ranklist untill and unless you are done with solving problems, wastes 3/4 minuts 
# Donot treat CP as a placement thing, love it and enjoy it, you will succeed for sure.  
# Any doubts or want to have a talk, contact https://www.facebook.com/chaudhary.mayank


# ///==========Libraries, Constants and Functions=============///

import sys
from bisect import bisect_left,bisect_right,insort
from collections import deque,Counter
from math import gcd,sqrt,factorial,ceil,log10,log2,floor
from itertools import permutations
from heapq import heappush,heappop,heapify
inf = float("inf")
mod = 998244353
#sys.setrecursionlimit(10**9)


def factorial_p(n, p):
    ans = 1
    if n <= p // 2:
        for i in range(1, n + 1):
            ans = (ans * i) % p
    else:
        for i in range(1, p - n):
            ans = (ans * i) % p
        ans = pow(ans, p - 2, p)
        if n % 2 == 0:
            ans = p - ans
    return ans


def nCr_p(n, r, p):
    ans = 1
    while (n != 0) or (r != 0):
        a, b = n % p, r % p
        if a < b:
            return 0
        ans = (ans * factorial_p(a, p) * pow(factorial_p(b, p), p - 2, p) * pow(factorial_p(a - b, p), p - 2, p)) % p
        n //= p
        r //= p
    return ans

def prime_sieve(n):
    """returns a sieve of primes >= 5 and < n"""
    flag = n % 6 == 2
    sieve = bytearray((n // 3 + flag >> 3) + 1)
    for i in range(1, int(n**0.5) // 3 + 1):
        if not (sieve[i >> 3] >> (i & 7)) & 1:
            k = (3 * i + 1) | 1
            for j in range(k * k // 3, n // 3 + flag, 2 * k):
                sieve[j >> 3] |= 1 << (j & 7)
            for j in range(k * (k - 2 * (i & 1) + 4) // 3, n // 3 + flag, 2 * k):
                sieve[j >> 3] |= 1 << (j & 7)
    return sieve
def prime_list(n):                            #<----- You have to call me if you wanna get list of primes upto range n
    """returns a list of primes <= n"""
    res = []
    if n > 1:
        res.append(2)
    if n > 2:
        res.append(3)
    if n > 4:
        sieve = prime_sieve(n + 1)
        res.extend(3 * i + 1 | 1 for i in range(1, (n + 1) // 3 + (n % 6 == 1)) if not (sieve[i >> 3] >> (i & 7)) & 1)
    return res
    
def is_prime(n):
    """returns True if n is prime else False"""
    if n < 5 or n & 1 == 0 or n % 3 == 0:
        return 2 <= n <= 3
    s = ((n - 1) & (1 - n)).bit_length() - 1
    d = n >> s
    for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
        p = pow(a, d, n)
        if p == 1 or p == n - 1 or a % n == 0:
            continue
        for _ in range(s):
            p = (p * p) % n
            if p == n - 1:
                break
        else:
            return False
    return True

def all_factors(n):
    """returns a sorted list of all distinct factors of n"""
    small, large = [],[]
    for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
        if not n % i:
            small.append(i)
            large.append(n//i)
    large.reverse()
    small.extend(large)
    return small

def count(n):
    return ((n*(n+1))//2)

def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
# ///===========MAIN=============///
n=int(input())
result=all_factors(n)
length=len(result)
flag=0
for i in range(length//2):
    if result[i]>=5 and result[length-i-1]>=5:
        flag=1
        n=result[i];m=result[length-i-1]
        break
if flag==0:
    print(-1)
    exit()
mat=[]
word='aeiou'
word=word*(m//5)
z=m%5
word+=word[:z]
mat.append(word)
l=len(word)
for i in range(1,n):
    new_word=word[1:]
    new_word+=word[l%5]
    mat.append(new_word)
    word=new_word
print(''.join(mat))