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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cassert>
#include <queue>
#include <set>
#include <vector>
#include <cmath>
#include <bitset>
#include <functional>
#include <limits>
#include <map>
using namespace std;

typedef long long ll;
#define rep1(i, n) for(ll i = 1; i <=(ll)(n); i++)
#define rep0(i, n) for(ll i = 0; i <=(ll)(n); i++)

ll pow(ll a, ll n) { //compute a^n with o(logn)
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * a;
        a = a * a;
        n >>= 1;
    }
    return res;
}

ll pow(ll a, ll n, ll mod) { //compute a^n (mod mod) with o(logn)
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

const int MAX = 199;
const int MOD = 1000000007;
ll fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
ll COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

//順列計算
ll PER(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[n - k] % MOD) % MOD;
}

int main(){
    string s;
    cin >> s;
    if(s == "RRR"){
        cout << 3 << endl;
    }else if(s == "RRS" || s == "SRR"){
        cout << 2 << endl;
    }else if(s == "RSS" || s == "SSR" || s == "SRS" || s == "RSR"){
        cout << 1 << endl;
    }else{
        cout << 0 << endl;
    }
    return 0;
}