There is a frog staying to the left of the string $s = s_1 s_2 \ldots s_n$ 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 $t$ ($1 \le t \le 10^4$) — 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 $2 \cdot 10^5$ characters 'L' and 'R'.

It is guaranteed that the sum of lengths of strings over all test cases does not exceed $2 \cdot 10^5$ ($\sum |s| \le 2 \cdot 10^5$).


-----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:

[Image]

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.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 't = int(input())\nfor i in range(t):\n    s = input() + "R"\n    mx = -float(\'inf\')\n    cur = 0\n    while s.find(\'R\', cur) != -1:\n        nxt = s.find(\'R\', cur)\n        d = nxt - cur\n        if d > mx:\n            mx = d\n        cur = nxt + 1\n    print(mx + 1)\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13047