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: Accepted)

// #pragma GCC target("avx2")  // CPU 処理並列化
// #pragma GCC optimize("O3")  // CPU 処理並列化
// #pragma GCC optimize("unroll-loops")  // 条件処理の呼び出しを減らす
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
#include<numeric>
using namespace std;
const long long mod=998244353;
const long long inf=mod*mod;
const long long d2=(mod+1)/2;
const double EPS=1e-10;
const double INF=1e+10;
const double PI=acos(-1.0);
const int C_SIZE = 3121000;
long long fact[C_SIZE];
long long finv[C_SIZE];
long long inv[C_SIZE];
long long Comb(int a,int b){
 	if(a<b||b<0)return 0;
 	return fact[a]*finv[b]%mod*finv[a-b]%mod;
}
void init_C(int n){
	fact[0]=finv[0]=inv[1]=1;
	for(int i=2;i<n;i++){
		inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
	}
	for(int i=1;i<n;i++){
		fact[i]=fact[i-1]*i%mod;
		finv[i]=finv[i-1]*inv[i]%mod;
	}
}
long long pw(long long a,long long b){
	if(a<0LL)return 0;
	if(b<0LL)return 0;
	long long ret=1;
	while(b){
		if(b%2)ret=ret*a%mod;
		a=a*a%mod;
		b/=2;
	}
	return ret;
}
int ABS(int a){return max(a,-a);}
long long ABS(long long a){return max(a,-a);}
double ABS(double a){return max(a,-a);}
// ここから編集しろ
char in[110000];
int main(){
	scanf("%s",in);
	int n=strlen(in);
	int ret=0;
	int cur=0;
	for(int i=0;i<n;i++){
		if(in[i]=='A'||in[i]=='T'||in[i]=='G'||in[i]=='C')cur++;
		else cur=0;
		ret=max(ret,cur);
	}
	printf("%d\n",ret);
}