You are given a string s, consisting of small Latin letters. Let's denote the length of the string as |s|. The characters in the string are numbered starting from 1. 

Your task is to find out if it is possible to rearrange characters in string s so that for any prime number p ≤ |s| and for any integer i ranging from 1 to |s| / p (inclusive) the following condition was fulfilled sp = sp × i. If the answer is positive, find one way to rearrange the characters.

Input

The only line contains the initial string s, consisting of small Latin letters (1 ≤ |s| ≤ 1000).

Output

If it is possible to rearrange the characters in the string so that the above-mentioned conditions were fulfilled, then print in the first line "YES" (without the quotes) and print on the second line one of the possible resulting strings. If such permutation is impossible to perform, then print the single string "NO".

Examples

Input

abc


Output

YES
abc


Input

abcd


Output

NO


Input

xxxyxxx


Output

YES
xxxxxxy

Note

In the first sample any of the six possible strings will do: "abc", "acb", "bac", "bca", "cab" or "cba".

In the second sample no letter permutation will satisfy the condition at p = 2 (s2 = s4).

In the third test any string where character "y" doesn't occupy positions 2, 3, 4, 6 will be valid.
import java.util.*;
import java.io.*;
public class a {
    static long mod = 1000000007;
    public static void main(String[] args) throws IOException {
        // Scanner input = new Scanner(new File("input.txt"));
        // PrintWriter out = new PrintWriter(new File("output.txt"));
        input.init(System.in);
        PrintWriter out = new PrintWriter((System.out));
        
        String s = input.next();
        int n = s.length();
        boolean[] primes = sieve(n);
        int[] comp = new int[n];
        int[] freq = new int[26];
        for(int i = 0; i<n; i++) freq[s.charAt(i)-'a']++;
        int count2 = 0;
        for(int i = 0; i<n; i++)
        {
            if(i>1 && !primes[i+1] || primes[i+1] && (2*(i+1) <= n)) {comp[i] = 0; count2++;}
            else comp[i] = 1;
        }
        int max = 0, maxdex = 0;
        for(int i =0; i<26; i++)
        {
            if(freq[i] > max) max = freq[maxdex=i];
        }
        if(max >= count2)
        {
            out.println("YES");
            freq[maxdex] -= count2;
            for(int i = 0; i<n; i++)
            {
                if(comp[i] == 0) out.print((char)('a'+maxdex));
                else
                {
                    for(int j = 0; j<26; j++)
                    {
                        if(freq[j] > 0)
                        {
                            freq[j]--;
                            out.print((char)('a'+j));
                            break;
                        }
                    }
                }
            }
        }
        else out.println("NO");
        
        
        out.close();
    }
    static long pow(long x, long p) {
        if (p == 0)
            return 1;
        if ((p & 1) > 0) {
            return (x * pow(x, p - 1)) % mod;
        }
        long sqrt = pow(x, p / 2);
        return (sqrt * sqrt) % mod;
    }

    static class input {
        static BufferedReader reader;
        static StringTokenizer tokenizer;

        /** call this method to initialize reader for InputStream */
        static void init(InputStream input) {
            reader = new BufferedReader(new InputStreamReader(input));
            tokenizer = new StringTokenizer("");
        }

        /** get next word */
        static String next() throws IOException {
            while (!tokenizer.hasMoreTokens()) {
                // TODO add check for eof if necessary
                tokenizer = new StringTokenizer(reader.readLine());
            }
            return tokenizer.nextToken();
        }

        static int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        static double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        static long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        static String nextLine() throws IOException {
            return reader.readLine();
        }
    }
    public static boolean[] sieve(int n)
    {
        boolean[] res = new boolean[n+1];
        for(int i = 2; i<=n; i++) res[i] = true;
        for(int i = 2; i <= Math.sqrt(n); i++)
            if(res[i])
                for(long j = (long)i*i; j<=n; j+=i)
                    res[(int) j] = false;
        return res;
    }
    //GCD - O(log(max(p, q)))
    //res[0] is gcd of p and q
    //res[1] and res[2] are values used in extended Euclidean algorithm for inverse mod
    public static long[] gcd(long p, long q)
    {
        if(q==0)
            return new long[] {p, 1, 0};
        long[] vals = gcd(q, p%q);
        return new long[] {vals[0], vals[2], vals[1] - (p/q)*vals[2]};
    }
    //Inverse Mod - O(log(mod))
    //Returns A s.t. x*A = 1(MOD mod)
    public static long invmod(long x)
    {
        long[] vals = gcd(x, mod);
        if(vals[0]>1)
            return -1;
        if(vals[1]>0)
            return vals[1];
        return mod + vals[1];
    }
}

