Prime Permutation(思维好题 )

Prime Permutation

codeforces 123A

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 s_p = s_p × 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.

题意：长度为n的字符串s,问是否能将s重新排列
满足对任意一个素数p 都有 s[p]=s[p*i] [i=1,2..n/p] n<=1000.通俗一点，就是2 3 4 6...这些位置上的字母必须一样。我们发现：如果一个质数*2还小于等于给定字符串的总长度，那么这个质数的倍数和比它小的质数和它们的倍数的位置上的字母是一样的。

质数p,q满足p*q<=n 那么p*2，q*2<=n
若质数p*2>n 那么位置p可以放任意一个字符.
若质数p*2<=n 那么这些位置都会等于同一个字符.

所以标记这些放相同字符的位置然后看出现次数最多的字符能否放入即可.

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <queue>
  5 #include <vector>
  6 #include <algorithm>
  7 using namespace std;
  8 const int maxn=1010;
  9 char s[maxn];
 10 int flag;
 11 vector<int>v;
 12 char ans[maxn];
 13 int cnt[maxn],prime[maxn];
 14 struct node{
 15     int zimu;
 16     int num;
 17     friend bool operator < (node a,node b)
 18     {
 19         return a.num<b.num;
 20     }
 21 };
 22 void init()
 23 {
 24     for(int i=2;i<maxn;i++)
 25     {
 26         if(!prime[i])
 27         {
 28             v.push_back(i);
 29             for(int j=i+i;j<maxn;j+=i)
 30             {
 31                 prime[j]=1;
 32             }
 33         }
 34     }
 35 }
 36 int main()
 37 {
 38     scanf("%s",s+1);
 39     int len=strlen(s+1);
 40     init();
 41     memset(prime,0,sizeof(prime));
 42     for(int i=0;i<v.size();i++)
 43     {
 44         if(v[i]*2>len)
 45             break;
 46         for(int j=1;j*v[i]<=len;j++)
 47         {
 48             prime[j*v[i]]=1;
 49         }
 50     }
 51     int tot=0;
 52     for(int i=1;i<=len;i++)
 53     {
 54         cnt[s[i]-'a']++;
 55         if(prime[i])
 56         {
 57             tot++;
 58         }
 59     }
 60     priority_queue<node>q;
 61     for(int i=0;i<26;i++)
 62     {
 63             node x;
 64             x.num=cnt[i];
 65             x.zimu=i;
 66             q.push(x);
 67     }
 68     node head=q.top();
 69     if(head.num<tot)
 70     {
 71         puts("NO");
 72         return 0;
 73     }
 74     puts("YES");
 75     int res=0;
 76     for(int i=1;i<=len;i++)
 77     {
 78         if(prime[i])
 79         {
 80             res++;
 81             ans[i]=q.top().zimu+'a';
 82          
 83         }
 84     }
 85     head=q.top();
 86     q.pop();
 87     head.num-=res;
 88     if(head.num)
 89     {
 90         q.push(head);
 91     }
 92     for(int i=1;i<=len;i++)
 93     {
 94         if(!prime[i])
 95         {
 96             head=q.top();
 97             ans[i]=head.zimu+'a';
 98             head.num--;
 99             q.pop();
100             if(head.num)
101                 q.push(head);
102         }
103     }
104     for(int i=1;i<=len;i++)
105         printf("%c",ans[i]);
106     printf("
");
107 }