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.
import java.io.*;
import java.util.*;

public class Main {
    private boolean eof;
    private BufferedReader br;
    private StringTokenizer st;
    private PrintWriter out;

    public static void main(String[] args) throws IOException {
        new Main().run();
    }

    private String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return "-1";
            }
        }
        return st.nextToken();
    }

    private int nextInt() {
        return Integer.parseInt(nextToken());
    }

    private long nextLong() {
        return Long.parseLong(nextToken());
    }

    private double nextDouble() {
        return Double.parseDouble(nextToken());
    }

    private String nextLine() throws IOException {
        return br.readLine();
    }


    private void run() throws IOException {
        InputStream input = System.in;
        PrintStream output = System.out;
        try {
            File f = new File("a.in");
            if (f.exists() && f.canRead()) {
                input = new FileInputStream(f);
                output = new PrintStream("a.out");
            }
        } catch (Throwable ignored) {
        }
        br = new BufferedReader(new InputStreamReader(input));
        out = new PrintWriter(new PrintWriter(output));
        solve();
        br.close();
        out.close();
    }

    private void solve() {
        int n = nextInt();
        String word = "aeiou";

        for (int i = 5; i < n; i++) {
            if (n % i == 0 && n / i >= 5) {
                int rows = i;
                int colums = n / i;

                for (int p = 0; p < rows; p++) {
                    for (int q = 0; q < colums; q++) {
                        out.print(word.charAt((p + q) % 5));
                    }
                }


                return;
            }
        }

        out.print(-1);
    }


    private static Random rnd = new Random();

    //Дерево поиска по ключу, балансится за счет случайных приоритетов
    //Поддерживает сумму(мин,макс) элементов в поддереве и нахождение суммы(мин,макс)на заданном отрезке.
    //
    public static class Node {
        public long key;
        public int prior;
        public long sum;
        public long min;
        public long max;
        public int cnt;
        public int level; //Номер уровня снизу вверх
        public Node left, right;

        Node(long key) {
            this.key = key;
            this.prior = rnd.nextInt();
            left = right = null;
            this.sum = key;
            this.min = key;
            this.max = key;
            this.cnt = 1;
            this.level = 1;
        }
    }

    public static long getMin(Node t) {
        return t == null ? Integer.MAX_VALUE : t.min;
    }

    public static long getMax(Node t) {
        return t == null ? Integer.MIN_VALUE : t.max;
    }

    public static long getSum(Node t) {
        return t == null ? 0 : t.sum;
    }

    public static int getCnt(Node t) {
        return t == null ? 0 : t.cnt;
    }

    public static int getLevel(Node t) {
        return t == null ? 0 : t.level;
    }

    public static void update(Node t) {
        if (t != null) {
            t.sum = t.key + getSum(t.left) + getSum(t.right);
            t.min = Math.min(t.key, getMin(t.left));
            t.max = Math.max(t.key, getMax(t.right));
            t.cnt = 1 + getCnt(t.left) + getCnt(t.right);
            t.level = 1 + getLevel(t.left);
        }
    }

    public static Pair<Node, Node> split(Node t, long key) {
        if (t == null) return new Pair<>(null, null);
        if (key > t.key) {
            Pair<Node, Node> p = split(t.right, key);
            t.right = p.first;
            update(t);
            return new Pair<>(t, p.second);
        } else {
            Pair<Node, Node> p = split(t.left, key);
            t.left = p.second;
            update(t);
            return new Pair<>(p.first, t);
        }
    }

    //SPLIT ПО НЕЯВНОМУ КЛЮЧУ
    //add for root = 0
    public static Pair<Node, Node> split(Node t, long key, int add) {
        if (t == null) return new Pair<>(null, null);
        int curKey = add + getCnt(t.left);
        if (key > curKey) { //идем в правое поддерево
            Pair<Node, Node> p = split(t.right, key, curKey + 1);
            t.right = p.first;
            update(t);
            return new Pair<>(t, p.second);
        } else { //идем в левое поддерево
            Pair<Node, Node> p = split(t.left, key, add);
            t.left = p.second;
            update(t);
            return new Pair<>(p.first, t);
        }
    }

    public static Node merge(Node t1, Node t2) {
        if (t1 == null) return t2;
        if (t2 == null) return t1;
        if (t1.prior > t2.prior) {
            t1.right = merge(t1.right, t2);
            update(t1);
            return t1;
        } else {
            t2.left = merge(t1, t2.left);
            update(t2);
            return t2;
        }
    }

//    public static Node insert(Node t, Node newNode) {
//        if (t == null) {
//            t = newNode;
//        } else if (newNode.prior > t.prior) {
//            Pair<Node, Node> p = split(t, newNode.key);
//            newNode.left = p.first;
//            newNode.right = p.second;
//            t = newNode;
//        } else {
//            if (newNode.key < t.key) {
//                t.left = insert(t.left, newNode);
//            } else {
//                t.right = insert(t.right, newNode);
//            }
//        }
//        update(t);
//        return t;
//    }

    //Еще один insert, но вроде не такой быстрый
    public static Node insert(Node t, Node newNode) {
        if (t == null) {
            t = newNode;
        } else {
            Pair<Node, Node> p = split(t, newNode.key);
            t = merge(p.first, newNode);
            t = merge(t, p.second);
        }
        update(t);
        return t;
    }

    public static Node remove(Node t, long key) {
        if (t == null) return null;
        if (t.key == key) {
            t = merge(t.left, t.right);
        } else {
            if (key < t.key) {
                t.left = remove(t.left, key);
            } else {
                t.right = remove(t.right, key);
            }
        }
        update(t);
        return t;
    }

    public static Node find(Node t, long key) {
        if (t == null) return null;
        if (t.key == key) return t;
        if (t.key < key) {
            return find(t.right, key);
        } else {
            return find(t.left, key);
        }
    }

    /**
     * Вырезает отрезок от i до j и вставляет в позицию k нового массива
     * (при k = 0 вставляет вначало)
     *
     * @param t Корень дерева
     * @param i Начало отрезка (включительно)
     * @param j Конец отрезка (включительно)
     * @param k Позиция вставки
     * @return Новое дерево
     */
    public static Node replaceSegment(Node t, int i, int j, int k) {
        //Вырезаем отрезок и склеиваем
        Pair<Node, Node> split1 = split(t, i, 0);
        Pair<Node, Node> split2 = split(split1.second, j - i + 1, 0);
        Node merge1 = merge(split1.first, split2.second);
        Node removedSegment = split2.first;

//      out.println("Array without removed:" + getString(merge1));
//      out.println("Removed Segment:" + getString(removedSegment));

        //разделяем и вставляем
        Pair<Node, Node> split3 = split(merge1, k, 0);
        Node merge2 = merge(split3.first, removedSegment);
        Node merge3 = merge(merge2, split3.second);

        return merge3;
    }

    public static long getSum(Node t, long l, long r) {
        Pair<Node, Node> p1 = split(t, l);
        Pair<Node, Node> p2 = split(p1.second, r + 1);
        long ans = p2.first == null ? 0 : p2.first.sum;
        t = merge(p1.first, p2.first);
        t = merge(t, p2.second);
        return ans;
    }

    public static StringBuilder getKeysString(Node t) {
        if (t == null) return new StringBuilder();
        StringBuilder ans = new StringBuilder();
        if (t.left != null) ans.append(getKeysString(t.left));
        ans.append(t.key);
        ans.append(" ");
        if (t.right != null) ans.append(getKeysString(t.right));
        return ans;
    }

    public static class Pair<F, S> {
        public F first;
        public S second;

        Pair(F first, S second) {
            this.first = first;
            this.second = second;
        }

//        @Override
//        public int compareTo(Object o) {
//            return first.compareTo(((Pair)o).first);
//        }

        @Override
        public boolean equals(Object o) {
            if (!(o instanceof Main.Pair)) {
                return false;
            }
            Main.Pair<?, ?> p = (Main.Pair<?, ?>) o;
            return Objects.equals(p.first, first) && Objects.equals(p.second, second);
        }

        @Override
        public int hashCode() {
            return (first == null ? 0 : first.hashCode()) ^ (second == null ? 0 : second.hashCode());
        }
    }
}
