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)

#include<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<n;++i)
#include<string>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<int, int>P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
//max=({});
//条件式が真ならwhileの中身を回し続ける
//printf("%d\n", ans);

//pairの入力
//vector<pair<ll, ll>>work(n);
//rep(i, n) {
//	ll a, b;
//	cin >> a >> b;
//	work[i] = make_pair(a, b);
//for(auto p:mp)(mapの探索)
//printf("%.10f\n",なんちゃら)

int g[15][15];
const int INF = 1001001001;
const int dx[4] = { -1,0,1,0 };
const int dy[4] = { 0,-1,0,1 };

//最大公約数
ll gcd(ll x, ll y) {
	return y ? gcd(y, x % y) : x;
}

ll lcm(ll x, ll y) {
	if (x == 0 || y == 0)return 0;
	return (x / gcd(x, y) * y);
}

//素因数分解
map<ll, ll>ins;
void facterize(ll n) {
	ll a = 2;
	while (n >= a * a) {
		if (n % a == 0) {
			ins[a]++;
			ins[a] = ins[a] % 1000000007;
			n /= a;
		}
		else {
			a++;
		}
	}
	ins[n]++;
}

vector<ll>relation[20], money(20, 1);
char s[110][110];
int dp[110][110];

int keta(int x) {
	int y = 0;
	while (x > 0) {
		y++;
		x /= 10;
	}
	return y;
}

int main() {
	string s;
	cin >> s;
	int n = s.length();
	int ans = 0;
	rep(i, n) {
		int res = 0;
		for (int j = i;j < n;++j) {
			if (s[j] != 'A' && s[j] != 'C' && s[j] != 'G' && s[j] != 'T') {
				break;
			}
			res++;
			ans = max(ans, res);
		}
	}
	cout << ans << endl;
	return 0;
}