Problem: The only difference between problems C1 and C2 is that all values in input of problem C1 are distinct (this condition may be false for problem C2).

You are given a sequence a consisting of n integers. All these integers are distinct, each value from 1 to n appears in the sequence exactly once.

You are making a sequence of moves. During each move you must take either the leftmost element of the sequence or the rightmost element of the sequence, write it down and remove it from the sequence. Your task is to write down a strictly increasing sequence, and among all such sequences you should take the longest (the length of the sequence is the number of elements in it).

For example, for the sequence [2, 1, 5, 4, 3] the answer is 4 (you take 2 and the sequence becomes [1, 5, 4, 3], then you take the rightmost element 3 and the sequence becomes [1, 5, 4], then you take 4 and the sequence becomes [1, 5] and then you take 5 and the sequence becomes [1], the obtained increasing sequence is [2, 3, 4, 5]).

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n), where a_i is the i-th element of a. All these integers are pairwise distinct.

Output

In the first line of the output print k — the maximum number of elements in a strictly increasing sequence you can obtain.

In the second line print a string s of length k, where the j-th character of this string s_j should be 'L' if you take the leftmost element during the j-th move and 'R' otherwise. If there are multiple answers, you can print any.

Examples

Input


5
2 1 5 4 3


Output


4
LRRR


Input


7
1 3 5 6 7 4 2


Output


7
LRLRLLL


Input


3
1 2 3


Output


3
LLL


Input


4
1 2 4 3


Output


4
LLRL

Note

The first example is described in the problem statement.

Solution (in PYTHON3):

Solution:
import sys 
# sys.setrecursionlimit(10**6) 
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
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 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

"""*******************************************************"""
def main():
    n=inin()
    a=ain()
    j=0
    k=n-1
    f=0
    s=""
    p=0
    while(k>=j):
        # print(f,j,k,a[j],a[k])
        if(a[j]>f and a[k]>f):
            if(a[j]==a[k]):
                p=1
                ss=s[:]
                sss=s[:]
                ss+="L"
                sss+="R"
                for x in range(j+1,k+1):
                    if(a[x]>a[x-1]):
                        ss+="L"
                    else:
                        break
                for x in range(k-1,j,-1):
                    if(a[x+1]<a[x]):
                        sss+="R"
                    else:
                        break
                break
            if(a[j]<a[k]):
                s+="L"
                f=a[j]
                j+=1
            else:
                s+="R"
                f=a[k]
                k-=1
        elif(a[j]>f):
            s+="L"
            f=a[j]
            j+=1
        elif(a[k]>f):
            s+="R"
            f=a[k]
            k-=1
        else:
            break
    if(p==1):
        if(len(ss)<len(sss)):
            print(len(sss))
            print(sss)
        else:
            print(len(ss))
            print(ss)
    else:
        print(len(s))
        print(s)
######## 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 different 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'0' [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'0' [0]:
        A.append(sign*numb)
    return A
 
if __name__== "__main__":
  main()