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 <bits/stdc++.h>

using namespace std;

typedef long long int ll;
typedef long long int lli;
typedef unsigned long long int ull;
typedef long double ld;
typedef string str;

typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<ld> vd;
typedef vector<str> vs;
typedef vector<ll> vll;
typedef vector<pair<ll, ll>> vpll;
typedef vector<vector<ll>> vvll;
typedef vector<vector<char>> vvc;
typedef vector<vector<str>> vvs;

#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define rrep(i, n) for (ll i = 1; i <= (n); ++i)
#define fin(ans) cout << (ans) << endl;
#define mp(p, q) make_pair(p, q)
//３つ以上の値をセットに(3より多いときはカッコ内を増やす)\
sort -> p -> q -> r
#define mt(p, q, r) make_tuple(p, q, r)
#define all(a) a.begin(), a.end()

template <typename T>
void V(T &n)
{
  for (ll i = 0; i < n.size(); i++)
  {
    cin >> n[i];
  }
}

template <typename T, typename U>
void V2(T &n, U &m)
{
  for (ll i = 0; i < n.size(); i++)
  {
    cin >> n[i] >> m[i];
  }
}

template <typename T>
void Sort(T &n, ll a = 0, ll b = 0)
{
  if (b == 0)
  {
    return (sort(n.begin() + a, n.end()));
  }
  else
  {
    return (sort(n.begin() + a, n.begin() + b));
  }
}

//配列を反転
template <typename T>
void Reve(T &n, ll a = 0, ll b = 0)
{
  if (b == 0)
  {
    return (reverse(n.begin() + a, n.end()));
  }
  else
  {
    return (reverse(n.begin() + a, n.begin() + b));
  }
}

template <typename T>
void Down(T &n, ll a = 0, ll b = 0)
{
  if (b == 0)
  {
    return (sort(n.begin() + a, n.end(), greater<ll>()));
  }
  else
  {
    return (sort(n.begin() + a, n.begin() + b, greater<ll>()));
  }
}

  void Main()
  {
    str s; cin >> s;
    ll ans = 0, flag = 0;
    rep(i, s.size())
    {
      if(s[i] == 'A'||s[i] == 'C'||s[i] == 'G'||s[i] == 'T')
      {
        flag++;
      }
      else
      {
        ans = max(ans, flag);
        flag = 0;
      }
    }
    fin(ans);
  }

  int main()
  {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(20); //高精度少数表示
    Main();
    return 0;
  }