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 (GCC 9.2.1) (judge verdict: Accepted)

/* ex 5_4
   nknm */

#include <stdio.h>
#include <string.h>

int main(void){
    

    char str[11], *ptr;      //必要な配列と整数を定義
    int i, count, max, *ptr_num;
    int num[11]={0};


    scanf("%s", str);     //文字列の入力

    ptr = str;      //ポインタを配列に先頭のアドレスへ
    ptr_num = num;

    count = 0;

    while(*ptr){
        if(*ptr == 'A' || *ptr == 'T' || *ptr == 'G' || *ptr == 'C'){
            count++;        //ATGCならカウントに1を足す
        }else{
            *ptr_num = count;       //ATGC以外のときにそれまでの連続するATGCの個数を他の配列に記録
            ptr_num++;      //記録用の配列のポインタを1つずらす
            count = 0;      //カウントを初期化
        }
        ptr++;      //次の文字を読む
    }
    *ptr_num = count;       //ATGCのいずれかで終わるときに最後に続くATGCの個数を記録
    
    max = num[0];       //記録した数の最大値を求める

    for(i = 0; i < 10; i++){
        if(num[i] < num[i+1]){
            max = num[i+1];
        }
    }

    printf("%d\n", max);

    return 0;
}