Problem Description

In Vivid’s school, there is a strange class(SC). In SC, the students’ names are very strange. They are in the same format: anbncn(a,b,c must not be the same with each other). For example studens whose names are“abc”,”ddppqq” are in SC, however studens whose names are “aaa”,“ab”,”ddppqqq” are not in SC.
Vivid makes friends with so many students, he wants to know who are in SC.

 

Input

There are multiple test cases (about 10), each case will give a string S which is the name of Vivid’s friend in a single line.
Please process to the end of file.

[Technical Specification]

1≤|S|≤10.

|S| indicates the length of S.

S only contains lowercase letter.

 

Output

For each case, output YES if Vivid’s friend is the student of SC, otherwise output NO.

 

Sample Input

 

Sample Output

 //#####################

 //Author:fraud

 //Blog: http://www.cnblogs.com/fraud/

 //#####################

 #include <iostream>

 #include <sstream>

 #include <ios>

 #include <iomanip>

 #include <functional>

 #include <algorithm>

 #include <vector>

 #include <string>

 #include <list>

 #include <queue>

 #include <deque>

 #include <stack>

 #include <set>

 #include <map>

 #include <cstdio>

 #include <cstdlib>

 #include <cmath>

 #include <cstring>

 #include <climits>

 #include <cctype>

 using namespace std;

 #define XINF INT_MAX

 #define INF 0x3FFFFFFF

 #define MP(X,Y) make_pair(X,Y)

 #define PB(X) push_back(X)

 #define REP(X,N) for(int X=0;X<N;X++)

 #define REP2(X,L,R) for(int X=L;X<=R;X++)

 #define DEP(X,R,L) for(int X=R;X>=L;X--)

 #define CLR(A,X) memset(A,X,sizeof(A))

 #define IT iterator

 typedef long long ll;

 typedef pair<int,int> PII;

 typedef vector<PII> VII;

 typedef vector<int> VI;

 int vis[];

 int num[];

 int main()

 {

     ios::sync_with_stdio(false);

     string s;

     while(cin>>s){

         int len=s.length();

         CLR(vis,);

         CLR(num,);

         int ans=;

         int cs=;

         int k=len/;

         if(k*!=len)ans=;if(ans){

         char str=s[];

         for(int i=;i<k;i++){

             if(s[i]!=str)ans=;

         }

         str=s[k];

         if(s[k]==s[k-])ans=;

         for(int i=k;i<len-k;i++){

             if(s[i]!=str)ans=;

         }

         str=s[len-k];

         if(str==s[len-k-])ans=;

         if(str==s[])ans=;

         for(int i=len-k;i<len;i++){

             if(s[i]!=str)ans=;

         }

         }

         if(ans)cout<<"YES"<<endl;

         else cout<<"NO"<<endl;

     }

     return ;

 }