#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) for(int i = 0; i < (n); i++)
#define repp(i, l, r) for(int i = (l); i < (r); i++)
#define per(i, n) for(int i = ((n)-1); i >= 0; i--)
#define perr(i, l, r) for(int i = ((r)-1); i >= (l); i--)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define IINF 1000000000
#define LINF 1000000000000000000
#define SP <<" "<<
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define CFS cin.tie(0);ios::sync_with_stdio(false)

typedef long long LL;
typedef long double LD;

int main(){
  string s;
  cin >> s;
  if(s.length()<3){
    cout << "No" << endl;
    return 0;
  }
  while(1){
    bool ok=false;
    rep(i,s.length()-2){
      if(s.substr(i,3)=="ABC"){
        s[i]=s[i+1]=s[i+2]='?';
        ok=true;
      }
    }
    if(!ok){
      cout << "No" << endl;
      return 0;
    }
    if(s=="???"){
      cout << "Yes" << endl;
      return 0;
    }
    vector<bool> f(3,false);
    rep(i,s.length()){
      if(s[i]!='?') f[s[i]-'A']=true;
    }
    char c='?';
    rep(i,3){
      if(!f[i]){
        if(c=='?'){
          c='A'+i;
        }else{
          cout << "No" << endl;
          return 0;
        }
      }
    }
    string ns="";
    rep(i,s.length()){
      if(s[i]=='?'){
        ns.push_back(c);
        i+=2;
      }else{
        ns.push_back(s[i]);
      }
    }
    s=ns;
  }
  return 0;
}


### Prompt

Develop a solution in CPP to the problem described below:

ABC Gene

There is a gene sequence represented by the string `ABC`. You can rewrite this gene sequence by performing the following operations several times.

* Choose one of the letters `A`,` B`, `C`. Let this be x. Replace all x in the gene sequence with `ABC` at the same time.



Given a string S consisting only of `A`,` B`, and `C`. Determine if the gene sequence can be matched to S.

Constraints

* 1 ≤ | S | ≤ 5,000
* S consists only of `A`,` B`, and `C`.



Input Format

Input is given from standard input in the following format.


S


Output Format

Output `Yes` if the gene sequence can be matched to S, and` No` if it cannot be matched.

Sample Input 1


ABC


Sample Output 1


Yes


The gene sequence is `ABC` from the beginning.

Sample Input 2


AABCC


Sample Output 2


Yes


If you select `B` and perform the operation, it becomes` ABC` → `AABCC`.

Sample Input 3


AABCABC


Sample Output 3


No


For example, even if you select `C` and perform an operation, it does not change from` AABCC` to `AABCABC`. Since all `C`s are replaced with` ABC` at the same time, the actual result is `AABCC` →` AABABCABC`.





Example

Input

ABC


Output

Yes

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) for(int i = 0; i < (n); i++)
#define repp(i, l, r) for(int i = (l); i < (r); i++)
#define per(i, n) for(int i = ((n)-1); i >= 0; i--)
#define perr(i, l, r) for(int i = ((r)-1); i >= (l); i--)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define IINF 1000000000
#define LINF 1000000000000000000
#define SP <<" "<<
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define CFS cin.tie(0);ios::sync_with_stdio(false)

typedef long long LL;
typedef long double LD;

int main(){
  string s;
  cin >> s;
  if(s.length()<3){
    cout << "No" << endl;
    return 0;
  }
  while(1){
    bool ok=false;
    rep(i,s.length()-2){
      if(s.substr(i,3)=="ABC"){
        s[i]=s[i+1]=s[i+2]='?';
        ok=true;
      }
    }
    if(!ok){
      cout << "No" << endl;
      return 0;
    }
    if(s=="???"){
      cout << "Yes" << endl;
      return 0;
    }
    vector<bool> f(3,false);
    rep(i,s.length()){
      if(s[i]!='?') f[s[i]-'A']=true;
    }
    char c='?';
    rep(i,3){
      if(!f[i]){
        if(c=='?'){
          c='A'+i;
        }else{
          cout << "No" << endl;
          return 0;
        }
      }
    }
    string ns="";
    rep(i,s.length()){
      if(s[i]=='?'){
        ns.push_back(c);
        i+=2;
      }else{
        ns.push_back(s[i]);
      }
    }
    s=ns;
  }
  return 0;
}


```