Problem:
Score : 200 points

Problem Statement

You are given a string S consisting of uppercase English letters. Find the length of the longest ACGT string that is a substring (see Notes) of S.

Here, a ACGT string is a string that contains no characters other than A, C, G and T.

Notes

A substring of a string T is a string obtained by removing zero or more characters from the beginning and the end of T.

For example, the substrings of ATCODER include TCO, AT, CODER, ATCODER and (the empty string), but not AC.

Constraints

S is a string of length between 1 and 10 (inclusive).

Each character in S is an uppercase English letter.

Input

Input is given from Standard Input in the following format:

S

Output

Print the length of the longest ACGT string that is a substring of S.

Sample Input 1

ATCODER

Sample Output 1

3

Among the ACGT strings that are substrings of ATCODER, the longest one is ATC.

Sample Input 2

HATAGAYA

Sample Output 2

5

Among the ACGT strings that are substrings of HATAGAYA, the longest one is ATAGA.

Sample Input 3

SHINJUKU

Sample Output 3

0

Among the ACGT strings that are substrings of SHINJUKU, the longest one is (the empty string).

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//INCLUDE
//------------------------------------------
#include <bits/stdc++.h>
//DEFINE
//------------------------------------------
#define ll long long
#define ld long double
#define ALLv(a)  (a).begin(),(a).end()
#define ALL(a,n)  (a),(a)+n
#define vi vector<long long>
#define vd vector<long double>
#define vs vector<string>
//CONST
//------------------------------------------
#define INF 1010000000000000017LL
#define MOD 1000000007LL
#define EPS 1e-12
#define PI 3.14159265358979323846
//REPEAT
//------------------------------------------
#define   FOR(i,m,n) for(ll (i)=(m);   (i)<(n);  (i)++)
#define   REP(i,n)   for(ll (i)=0;     (i)<(n);  (i)++)
#define  REPS(i,x)   for(ll (i)=1;     (i)<=(x); (i)++)
#define  RREP(i,x)   for(ll (i)=(x)-1; (i)>=0;   (i)--)
#define RREPS(i,x)   for(ll (i)=(x);   (i)> 0;   (i)--)
#define  WREP(i,in,j,jn) REP(i,in)REP(j,jn)
//-----------------------------------------
#define ifYN(x) cout<<(x?"Yes":"No")<<"\n" 
#define dcml(n) fixed<<setprecision(n)

 
ll GCD(ll a,ll b) { return b ? GCD(b,a%b) : a; } 
ll LCM(ll a,ll b) { return a/GCD(a,b)*b; } 
ll Factorial(ll n){ ll m=1;while(n>=1)m*=n--;return m; } 
ll modPow(ll x,ll n,ll mod=MOD){ 
    return (!n?1:!n%2?modPow(x*x,n/2,mod)%mod:x%mod*modPow(x,n-1,mod)%mod); 
} 
ll GetDigit(ll num){ 
    ll digit=0; 
    while(num!=0){num/=10;digit++;} 
    return digit; 
} 


using namespace std;
//-------------------------------------------

int main(void){
  cin.tie(0);
  ios::sync_with_stdio(false);
  string S;cin>>S;
  ll ans=0;ll cnt=0;
  REP(i,S.size()){
    if(S[i]=='A'||S[i]=='C'||S[i]=='G'||S[i]=='T'){
      cnt++;
    }else{
      ans=max(ans,cnt);
      cnt=0;
    }
  }
  cout<<ans;
  return 0;
}