题干：

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.

题目大意：

给你一个字符串(len<=1000)，让你找到它的另一个排列，使得任意一个素数的倍数的位置上的字符相同。

解题报告：

先说正解：不难发现，除了1和大于len/2的素数  其他都需要是同一个数。

证明如下：对于小于等于len/2的数，如果他是合数，那肯定要和2位置上的字符相同，如果他是素数，假设为p，那他要和2*p的位置相同，而2*p的位置和2的相同，所以也要相同。对于大于len/2的数，显然。

再说另一种做法：并查集，最直观的想法，但是实现的时候有个问题，见下面的代码，如果加上getf那一行循环，对于len==20左右的时候就GG，直接在下面cmp1那个排序中死循环，不知道为什么。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
char s[MAX];
bool is[MAX];
int n;
struct Node {char ch;int cnt;bool operator<(const Node & b) const {return cnt < b.cnt;}
} R[MAX];
struct NNode {int f,num,id;
} W[MAX],WW[MAX];
bool cmp1(NNode a,NNode b) {if(a.f == b.f) return 1;else return WW[a.f].num < WW[b.f].num;
}
bool cmp2(NNode a,NNode b) {return a.id < b.id;
}
int getf(int v) {return W[v].f == v ? v : W[v].f = getf(W[v].f);
}
void merge(int u,int v) {int t1 = getf(u),t2=getf(v);if(t1 == t2) return;W[t2].f = t1;W[t1].num += W[t2].num;
}
void init() {memset(is,1,sizeof is);is[0]=is[1]=0;for(int i = 2; i<=n; i++) {if(is[i] == 0) continue;for(int j = i+i; j<=n; j+=i) {is[j] = 0;merge(i,j);}}
}
char ans[MAX];
int main()
{	for(char i = 'a'; i<='z'; i++) R[i].ch=i,R[i].cnt=0; scanf("%s",s+1); n = strlen(s+1);for(int i = 1; i<=n; i++) W[i].f=i,W[i].num=1,W[i].id=i,R[s[i]].cnt++;	init();sort(R+'a',R+'z'+1); 
//	for(int i = 1; i<=n; i++) getf(i);for(int i = 1; i<=n; i++) WW[i]=W[i];sort(W+1,W+n+1,cmp1);int cur = 'a',flag=1;//注意cur只代表ascll码，不代表此时字符就是a字符，因为你已经排序了 for(int i = 1; i<=n; i++) {while(R[cur].cnt == 0 && cur <= 'z') cur++;if(cur == 'z'+1) {flag=0;break;	}ans[W[i].id] = R[cur].ch;R[cur].cnt--;}sort(W+1,W+n+1,cmp2);//排序还原回原序列 使得和getf可以对应上for(int i = 1; i<=n; i++) {if(ans[i] != ans[getf(i)]) flag = 0; }if(flag == 0) {puts("NO");return 0;}puts("YES");printf("%s",ans+1);return 0 ;
}