FrogJumps.cpp

/*
Code Forces #627 C
Frog Jumps

time limit per test: 2 seconds
memory limit per test: 256 megabytes
input: standard input
output: standard output

There is a frog staying to the left of the string s=s1s2…sn consisting of n characters
(to be more precise, the frog initially stays at the cell 0). Each character of s is either 'L' or 'R'.
It means that if the frog is staying at the i-th cell and the i-th character is 'L',
the frog can jump only to the left. If the frog is staying at the i-th cell and the i-th character is 'R',
the frog can jump only to the right. The frog can jump only to the right from the cell 0.

Note that the frog can jump into the same cell twice and can perform as many jumps as it needs.

The frog wants to reach the n+1-th cell. The frog chooses some positive integer value d before
the first jump (and cannot change it later) and jumps by no more than d cells at once.
I.e. if the i-th character is 'L' then the frog can jump to any cell in a range [max(0,i−d);i−1],
and if the i-th character is 'R' then the frog can jump to any cell in a range [i+1;min(n+1;i+d)].

The frog doesn't want to jump far, so your task is to find the minimum possible value of d
such that the frog can reach the cell n+1 from the cell 0 if it can jump by no more
than d cells at once. It is guaranteed that it is always possible to reach n+1 from 0.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤104) — the number of test cases.

The next t lines describe test cases. The i-th test case is described as a string s consisting
of at least 1 and at most 2⋅105 characters 'L' and 'R'.

It is guaranteed that the sum of lengths of strings over all test cases does not exceed 2⋅105 (∑|s|≤2⋅105).

Output
For each test case, print the answer — the minimum possible value of d such that the frog can
reach the cell n+1 from the cell 0 if it jumps by no more than d at once.

Example:

input

6
LRLRRLL
L
LLR
RRRR
LLLLLL
R

output

3
2
3
1
7
1

Note
The picture describing the first test case of the example and one of the possible answers:

In the second test case of the example, the frog can only jump directly from 0 to n+1.

In the third test case of the example, the frog can choose d=3, jump to the cell 3 from the cell 0 and then to the cell 4 from the cell 3.

In the fourth test case of the example, the frog can choose d=1 and jump 5 times to the right.

In the fifth test case of the example, the frog can only jump directly from 0 to n+1.

In the sixth test case of the example, the frog can choose d=1 and jump 2 times to the right.

*/

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
 
#define ff              first
#define ss              second
#define int             long long
#define pb              push_back
#define mp              make_pair
#define pii             pair<int,int>
#define vi              vector<int>
#define mii             map<int,int>
#define pqb             priority_queue<int>
#define pqs             priority_queue<int,vi,greater<int> >
#define setbits(x)      __builtin_popcountll(x)
#define zrobits(x)      __builtin_ctzll(x)
#define mod             1000000007
#define inf             1e18
#define ps(x,y)         fixed<<setprecision(y)<<x
#define mk(arr,n,type)  type *arr=new type[n];
#define w(x)            int x; cin>>x; while(x--)
mt19937                 rng(chrono::steady_clock::now().time_since_epoch().count());
 
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
 
 
void c_p_c()
{
    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
}
 
int32_t main()
{
    c_p_c();
    w(t){
        string s;
        cin >> s;
        vi v;
        v.pb(0);
        
        int n = s.length();

        for(int i = 0; i < n; i++){
            if(s[i] == 'R'){
                v.pb(i + 1);
            }
        }

        v.pb(n + 1);

        int ans = 0;

        for(int i = 0; i < v.size() - 1; i++){
            ans = max(ans, v[i + 1] - v[i]);
        }

        cout << ans << '\n';
    }
    return 0;
}