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 int             long long
#define pb              push_back
//#define mp              make_pair
#define pii             pair<int,int>
//#define vec             vector<int>
#define mii             map<int,int>
#define pqb             priority_queue<int>
#define inf             1e18
#define ps(x,y)         fixed<<setprecision(y)<<x
#define no_of_test(x)   int x; cin>>x; while(x--)

const int mod = 1e9 + 7;
void very_fast()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
}
int gcd(int a, int b)
{
	if (b == 0) return a;
	return gcd(b, a % b);
}
int binary_exponentation(int a, int b)
{
	int ret = 1;
	while (b)
	{
		if (b & 1) ret = (ret * a) % mod;
		b >>= 1;
		a = (a * a) % mod;
	}
	return ret;
}
int32_t main()
{
	very_fast();

	{
		int a, b, c;
		cin >> a >> b >> c;
		int k = abs(a);
		int m = 1e18, n = 0;
		if (k > b * c)
		{
			cout << (k - b * c) << endl;
			return 0;
		}
		else
		{
			for (int i = 0; i < b; i++)
			{
				int w = k - c;
				k = k - c;
				m = min(m, abs(w));
				if (k - 0 <= c && (b + 1) % 2 == 0)
					break;
			}
			cout << abs(m) << endl;

		}

	}
	return 0;
}