codeforces 1324 C. Frog Jumps（贪心/二分）

There is a frog staying to the left of the string s=s1s2…sns=s1s2…sn consisting of nn characters (to be more precise, the frog initially stays at the cell 00 ). Each character of ss is either 'L' or 'R'. It means that if the frog is staying at the ii -th cell and the ii -th character is 'L', the frog can jump only to the left. If the frog is staying at the ii -th cell and the ii -th character is 'R', the frog can jump only to the right. The frog can jump only to the right from the cell 00 .

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+1n+1 -th cell. The frog chooses some positive integer value dd before the first jump (and cannot change it later) and jumps by no more than dd cells at once. I.e. if the ii -th character is 'L' then the frog can jump to any cell in a range [max(0,i−d);i−1][max(0,i−d);i−1] , and if the ii -th character is 'R' then the frog can jump to any cell in a range [i+1;min(n+1;i+d)][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 dd such that the frog can reach the cell n+1n+1 from the cell 00 if it can jump by no more than dd cells at once. It is guaranteed that it is always possible to reach n+1n+1 from 00 .

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤1041≤t≤104 ) — the number of test cases.

The next tt lines describe test cases. The ii -th test case is described as a string ss consisting of at least 11 and at most 2⋅1052⋅105 characters 'L' and 'R'.

It is guaranteed that the sum of lengths of strings over all test cases does not exceed 2⋅1052⋅105 (∑|s|≤2⋅105∑|s|≤2⋅105 ).

Output

For each test case, print the answer — the minimum possible value of dd such that the frog can reach the cell n+1n+1 from the cell 00 if it jumps by no more than dd at once.

Example

Input

6
LRLRRLL
L
LLR
RRRR
LLLLLL
R

Output

3
2
3
1
7
1
这个题队友用二分写的== 我当时想的贪心直接扫一遍就行...
懒得写了，放上队友的二分代码吧。

#include <bits/stdc++.h>

using namespace std;
const int maxn = 2 * 1e5 + 10;
char s[maxn];
int n;

bool check(int d) {
    int p = 0;
    while(p + d < n+1) {
        bool ok = false;
        for(int i = p + d; i >= p + 1; i--)
            if(s[i] == 'R') { p = i; ok = true; break; }
        if(ok == false) break;
    }
    return p + d >= n + 1;
}

int main() {
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while(t--) {
        memset(s, 0, sizeof s);
        cin >> s + 1;
        n = strlen(s + 1);
        int l = 0, r = n + 1;
        while(r - l > 1) {
            int mid = (l + r) / 2;
            if(check(mid)) r = mid;
            else l = mid;
        }
        printf("%d
", r);
    }
    return 0;
}