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.
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
    sys.stdin = open('input.txt', 'r')
    sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
MOD=1000000007
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]

#-------------------------code---------------------------#
# vsInput()

def primeN(n): 
    prime = [True for i in range(n+1)] 
    prime[0]=False
    prime[1]=False
    p=2
    while(p*p<=n): 
        if(prime[p]): 
            for i in range(p*p,n+1,p): 
                prime[i]=False
        p+=1
    return [p for p in range(n+1) if(prime[p])]



s=input()
n=len(s)
primes=primeN(n)
C=Counter(s)
ans=['' for i in range(n)]

have={}
for i in primes:
    have[i]=n//i

moreThanOne=[]
need=set()
for p in have:
    if(have[p]>1):
        moreThanOne.append(p)
        for i in range(p-1,n,p):
            need.add(i)
need=len(need)
# print(need,have)

ok=False
for i in C:
    if(C[i]>=need):
        ok=True
        ele=i
        C[i]-=need
        break


if(not ok):
    print("NO")
    exit()
print("YES")

for p in moreThanOne:
    for i in range(p-1,n,p):
        ans[i]=ele


rem=list(set(s))
ind=0

for i in range(n):
    if(ans[i]==''):
        while(C[rem[ind]]==0):
            ind+=1
        ans[i]=rem[ind]
        C[rem[ind]]-=1

print(*ans,sep="")


                

    
