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)

#include <bits/stdc++.h>
#include <chrono> 
 
using namespace std;
using namespace std::chrono; 
 
// #pragma GCC target ("avx2")
// #pragma GCC optimization ("O3")
// #pragma GCC optimization ("unroll-loops")
// #pragma optimization_level 3
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
#define f0r(a, b) for (long long a = 0; a < (b); ++a)
#define f1r(a, b, c) for (long long a = (b); a < (c); ++a)
#define f0rd(a, b) for (long long a = (b); a >= 0; --a)
#define f1rd(a, b, c) for (long long a = (b); a >= (c); --a)
#define ms(arr, v) memset(arr, v, sizeof(arr))
#define pb push_back
#define send {ios_base::sync_with_stdio(false);}
#define help {cin.tie(NULL); cout.tie(NULL);}
#define fix(prec) {cout << setprecision(prec) << fixed;}
#define mp make_pair
#define f first
#define s second
#define all(v) v.begin(), v.end()
#define getunique(v) {sort(all(v)); v.erase(unique(all(v)), v.end());}
#define readgraph(list, edges) for (int i = 0; i < edges; i++) {int a, b; cin >> a >> b; a--; b--; list[a].pb(b); list[b].pb(a);}
#define ai(a, n) for (int ele = 0; ele < n; ele++) cin >> a[ele];
#define ain(a, lb, rb) for (int ele = lb; ele <= rb; ele++) cin >> a[ele];
#define ao(a, n) {for (int ele = 0; ele < (n); ele++) { if (ele) cout << " "; cout << a[ele]; } cout << '\n';}
#define aout(a, lb, rb) {for (int ele = (lb); ele <= (rb); ele++) { if (ele > (lb)) cout << " "; cout << a[ele]; } cout << '\n';}
#define vsz(x) ((long long) x.size())
typedef long long ll;
typedef double ld;
typedef long double lld;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
 
template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v);
template<typename A, typename B> ostream& operator<<(ostream &cout, pair<A, B> const &p) { return cout << "(" << p.f << ", " << p.s << ")"; }
template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v) {
	cout << "["; for(int i = 0; i < v.size(); i++) {if (i) cout << ", "; cout << v[i];} return cout << "]";
}
template<typename A, typename B> istream& operator>>(istream& cin, pair<A, B> &p) {
	cin >> p.first;
	return cin >> p.second;
}
 
mt19937 rng(steady_clock::now().time_since_epoch().count());
// mt19937 rng(61378913);
/* usage - just do rng() */
 
void usaco(string filename) {
  // #pragma message("be careful, freopen may be wrong")
	freopen((filename + ".in").c_str(), "r", stdin);
	freopen((filename + ".out").c_str(), "w", stdout);
}
 
const lld pi = 3.14159265358979323846;
// const ll mod = 1000000007;
// const ll mod = 998244353;
// ll mod;



ll n, m, k, q, l, r, x, y, z;
ll a[1000005];
ll b[1000005];
ll c[1000005];
string s, t;
ll ans = 0;



void solve(int tc) {
	cin >> s;
	n = 3;
	
	ans = 0;
	ll r = 0;
	f0r(i, n) {
		if (s[i] == 'R') ++r;
		else r = 0;
		ans = max(ans, r);
	}
	
	cout << ans << '\n';
}

int main() {
	#ifdef galen_colin_local
		auto begin = std::chrono::high_resolution_clock::now();
	#endif
	
	send help

	// usaco("cowland");

	int tc = 1;
	// cin >> tc;
	for (int t = 0; t < tc; t++) solve(t);
	
	#ifdef galen_colin_local
		auto end = std::chrono::high_resolution_clock::now();
		cout << setprecision(4) << fixed;
		// cout << "Execution time: " << std::chrono::duration_cast<std::chrono::duration<double>>(end - begin).count() << " seconds" << endl;
	#endif
}