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>
using namespace std;
#define fastIO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define int long long int
#define fi first
#define se second
#define pub push_back
#define pi pair<int,int>
#define all(v) (v).begin(), (v).end()
#define rep(i, l, r) for(int i=(int)(l);i<(int)(r);i++)
#define repd(i, l, r) for (int i=(int)(l);i>=(int)(r);i--)
#define clrg(i, l, r) for(int i=(int)(l);i<(int)(r);i++)vis[i]=0,v[i].clear();
int power(int x, unsigned int y){int res = 1;while (y > 0){ if (y & 1){res = res*x;} y = y>>1;x = x*x;}return res;}
int powermod(int x, unsigned int y, int p){int res = 1;x = x % p;while (y > 0){if (y & 1){res = (res*x) % p;}y = y>>1; x = (x*x) % p;}return res;}
#define print2d(mat,n,m){for(int i=0;i<(int)(n);i++){for(int j=0;j<(m);j++){cout<<mat[i][j]<<" ";}cout<< endl;}}
#define clr(a,x) memset(a,x,sizeof(a))
#define rr(v) for(auto &val:v)
#define print(v) for (const auto itr : v){cout<<itr<<' ';}cout<<"\n";
#define ln length()
#define sz size()
#define mod 1000000007
#define elif else if
int32_t main(){

    string s;
     cin>>s;
    if(s[0]=='R' and s[1]=='R' and s[2]=='R'){
        cout<<3<<"\n";
    }
    elif(s[0]=='R' and s[1]=='R'){
        cout<<2<<"\n";
    }
    elif(s[1]=='R' and s[2]=='R'){
        cout<<2<<"\n";
    }
    elif(s[0]=='R' or s[1]=='R' or s[3]=='R')cout<<1<<"\n";
    else cout<<0<<"\n";
   return 0;
}
/*
Edge cases? n=1? a[i]<=0?
long vs int? 1LL? 64bits?
Re-read question. Is it as easy as it seems?
KEEP CORRECTING AND SUBMITTING!
*/