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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int INTMAX = numeric_limits<int>::max();
const long long LLMAX = numeric_limits<long long>::max();
void writeln() { cout << "\n"; }
template <class T>
inline void print(T&& a);
template <class T>
inline void priws(T&& a);
template <class T>
inline void read(T& a);
template <class... Args>
inline void readln(Args&... args) {
  (read(args), ...);
}
template <class H, class... T>
inline void writeln(H&& h, T&&... t) {
  priws(h);
  (print(t), ...);
  writeln();
}
template <class T>
inline void writeln_range(T f, T s) {
  if (f != s)
    for (auto i = f; i != s; ++i) writeln(*i);
}
vector<string> split(string& s, string d) {
  vector<string> v;
  size_t p = 0;
  while ((p = s.find(d)) != string::npos)
    v.push_back(s.substr(0, p)), s.erase(0, p + d.length());
  v.push_back(s);
  return v;
}
template <class... Args>
inline void err(string v, Args... args) {
  auto vv = split(v, ", ");
  auto it = vv.begin();
  (writeln(*it++, "=", args), ...);
}
void run() {
  int n;
  readln(n);
  vector<int> a(n);
  readln(a);
  string s;
  int i = 0;
  int j = n - 1;
  int last = 0;
  auto moveLeft = [&]() {
    last = a[i++];
    s.push_back('L');
  };
  auto moveRight = [&]() {
    last = a[j--];
    s.push_back('R');
  };
  while (i <= j) {
    if (a[i] > last && a[j] > last) {
      if (a[i] < a[j])
        moveLeft();
      else if (a[j] < a[i])
        moveRight();
      else {
        int I = i + 1;
        int J = j - 1;
        while (I < j && a[I] > a[I - 1]) ++I;
        while (J > i && a[J] > a[J + 1]) --J;
        int cl = I - i;
        int cr = j - J;
        if (cl > cr)
          for (int k = 0; k < (int)(cl); ++k) moveLeft();
        else
          for (int k = 0; k < (int)(cr); ++k) moveRight();
        break;
      }
    } else if (a[i] > last)
      moveLeft();
    else if (a[j] > last)
      moveRight();
    else
      break;
  }
  writeln(s.size());
  writeln(s);
}
int main() {
  ios_base::sync_with_stdio(false);
  run();
  return 0;
}
template <class T>
inline ostream& operator<<(ostream& os, vector<T> const& _a);
template <class T1, class T2>
inline istream& operator>>(istream& is, pair<T1, T2>& _a) {
  return is >> _a.first >> _a.second;
}
template <class T1, class T2>
inline ostream& operator<<(ostream& os, pair<T1, T2> const& _a) {
  return os << _a.first << " " << _a.second;
}
template <class T>
inline ostream& operator<<(ostream& os, vector<T> const& _a) {
  if (_a.size())
    os << _a[0];
  else
    os << "\n";
  for (int i = 1; i < (int)(_a.size()); ++i)
    os << "\n "[is_fundamental<T>::value] << _a[i];
  return os;
}
template <class T>
inline ostream& operator<<(ostream& os, valarray<T> const& _a) {
  if (_a.size())
    os << _a[0];
  else
    os << "\n";
  for (int i = 1; i < (int)(_a.size()); ++i)
    os << "\n "[is_fundamental<T>::value] << _a[i];
  return os;
}
template <class T>
inline istream& operator>>(istream& is, vector<T>& _a) {
  for (int i = 0; i < (int)(_a.size()); ++i) is >> _a[i];
  return is;
}
template <class T>
inline istream& operator>>(istream& is, valarray<T>& _a) {
  for (int i = 0; i < (int)(_a.size()); ++i) is >> _a[i];
  return is;
}
template <class T>
inline void print(T&& _a) {
  cout << " " << _a;
}
template <class T>
inline void priws(T&& _a) {
  cout << _a;
}
template <class T>
inline void read(T& _a) {
  cin >> _a;
}