Problem: Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string. 

The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.

The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.

Note that characters in the string in this problem are indexed from 1.

For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.

Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6

It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.

Input

The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.

Output

The first line should contain k (0≤ k ≤ 30) — the number of operations performed.

Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.

The length of the resulting palindrome must not exceed 10^6.

Examples

Input


abac


Output


2
R 2
R 5


Input


acccc


Output


2
L 4
L 2


Input


hannah


Output


0

Note

For the first example the following operations are performed:

abac → abacab → abacaba

The second sample performs the following operations: acccc → cccacccc → ccccacccc

The third example is already a palindrome so no operations are required.

Solution (in PYTHON):

Solution:
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
import sys
sys.setrecursionlimit(10**4)
# sys.stdin = open('inpy.txt', 'r')
# sys.stdout = open('outpy.txt', 'w')
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
i_m=9223372036854775807
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
import math  
def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x 
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l
prime=[]
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
def primeFactors(n): 
    a=[]
    # Print the number of two's that divide n 
    while n % 2 == 0: 
        a.append(2) 
        n = n // 2
          
    # n must be odd at this point 
    # so a skip of 2 ( i = i + 2) can be used 
    for i in range(3,int(math.sqrt(n))+1,2): 
          
        # while i divides n , print i ad divide n 
        while n % i== 0: 
            a.append(i) 
            n = n // i 
              
    # Condition if n is a prime 
    # number greater than 2 
    if n > 2: 
        a.append(n)
    return a

"""*******************************************************"""
def main():
    s=sin()
    n=len(s)
    print(3)
    print("R",n-1)
    print("L",n)
    print("L",2)
        

    
######## Python 2 and 3 footer by Pajenegod and c1729
 
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is yferent which makes it kind of a mess to use.
 
# So on cf, use PyPy2 for best string performance.
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'R' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'R' [0]:
        A.append(sign*numb)
    return A
 
# threading.Thread(target=main).start()
if __name__== "__main__":
  main()