Problem:
Score : 100 points

Problem Statement

We have weather records at AtCoder Town for some consecutive three days. A string of length 3, S, represents the records - if the i-th character is S, it means it was sunny on the i-th day; if that character is R, it means it was rainy on that day.

Find the maximum number of consecutive rainy days in this period.

Constraints

|S| = 3

Each character of S is S or R.

Input

Input is given from Standard Input in the following format:

S

Output

Print the maximum number of consecutive rainy days in the period.

Sample Input 1

RRS

Sample Output 1

2

We had rain on the 1-st and 2-nd days in the period. Here, the maximum number of consecutive rainy days is 2, so we should print 2.

Sample Input 2

SSS

Sample Output 2

0

It was sunny throughout the period. We had no rainy days, so we should print 0.

Sample Input 3

RSR

Sample Output 3

1

We had rain on the 1-st and 3-rd days - two "streaks" of one rainy day, so we should print 1.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
typedef long long ll;
const int INF = 1e9;
const int MOD = 1e9 + 7;
using namespace std;

//ループマクロ
#define FOR(i , a , b)  for(int i = (a) ; i < (b) ; i++)
#define REP(i , n)      for(int i = 0 ; i < n ; i++)
#define REPR(i , n)     for(int i = n ; 0 <= i ; i--)
#define ROP(i, n)       for(int i = 1 ; i <= n ; i++)

//入出力
#define SCOUT(x) cout << (x) << " "
#define VECCIN(x) for(auto& ele : (x) ) cin >> ele
#define VECOUT(x) for(auto& ele : (x) ) cout << ele << " "; cout << endl;

//その他
#define ALL(obj) (obj).begin() , (obj).end() 
#define SORT(obj) sort(ALL(obj))
#define GSORT(obj) sort(ALL(obj) , greater<int>())
#define P pair<int , int>
#define V vector<int>
#define VLL vector<long long>
#define M map<int, int>
#define S set<int>
#define PQ priority_queue<int>
#define PQG priority_queue<int , V , greater<int>>

int main(void) {

	ios::sync_with_stdio(false);	cin.tie(0);
    string s = "R", input;  cin >> input; s += input;

    int ans = 0, kp = 0;
    for(int i = 1 ; i <= 3 ; i++){
      if(s[i-1] == 'R' && s[i] == 'R')  kp += 1;
      else{
        ans = max(ans, kp);
        kp = 0;
      }
    }

    ans = max(ans, kp);
    cout << ans << endl;
	return 0;
}