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: Accepted)

/*      Author : Kaustubh Gangwar
        Institute : IIITA        */
#include <bits/stdc++.h>
using namespace std;
#define modo 1000000007
#define F first
#define S second 
#define pb push_back
#define gcd __gcd
#define vi vector<int>
#define vll vector<long long>
#define all(x) (x).begin(),(x).end()
#define f(i,a,b) for(int i=a;i<b;i++)
#define fe(i,a,b) for(int i=a;i<=b;i++)
#define CODING ios_base::sync_with_stdio(false); 
#define SAMRAT cin.tie(0),cout.tie(0);
#define mii map<int,int>
#define nl '\n'
typedef long long  ll;
typedef pair<int,int> pii;
 
ll bpow(ll a, ll b) {
    a=a; ll ans = 1;
    while(b){
        if (b & 1)
        ans = ans * a;
        a = a * a;
        b >>= 1;
    }
    return ans;
}

int32_t main()
{ 
   CODING SAMRAT
   int test=1;
   //cin>>test;
   while(test--)
   {
       string s; cin>>s;
       
       int ans=0;
       f(i,0,3)
       {   
           if(s[i]=='R')
           {   int cnt=1;
               f(j,i+1,3)
               {
                   if(s[j]!='R')
                   {
                       i=j-1;
                       break;
                   }
                   else
                   cnt++;
                   
                   if(j==2)
                   break;
               }
               
           ans=max(ans,cnt);
           }
       }
        cout<<ans<<nl;
    
    
   }

return 0;
}