CF--Increasing Subsequence--贪心

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 aa consisting of nn integers.

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 strictlyincreasing 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 [1,2,4,3,2][1,2,4,3,2] the answer is 44 (you take 11 and the sequence becomes [2,4,3,2][2,4,3,2], then you take the rightmost element 22 and the sequence becomes [2,4,3][2,4,3], then you take 33 and the sequence becomes [2,4][2,4] and then you take 44and the sequence becomes [2][2], the obtained increasing sequence is [1,2,3,4][1,2,3,4]).

Input

The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of elements in aa.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤2⋅1051≤ai≤2⋅105), where aiai is the ii-th element of aa.

Output

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

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

Examples

Input

5
1 2 4 3 2

Output

4
LRRR

Input

7
1 3 5 6 5 4 2

Output

6
LRLRRR

Input

3
2 2 2

Output

1
R

Input

4
1 2 4 3

Output

4
LLRR

Note

The first example is described in the problem statement.

 

每次都要从左边或者右边选一个数，求能组成的最长递增序列最长是多少，并输出每次是选左边还是右边。

思路：设立两个双指针，每次选择最小的且>上一个数的数，当两个数值相同是，我们找他们可以往左/右延续的最大值，最大值大者先选。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=200000+66;
const ll mod=1e9+7;
int q,n;
int a[maxn];
bool check(int l,int r,int m)
{
    if(l==r)
        return false;
    if(a[l]<a[r]&&a[l]>m)
        return true;
    else if(a[l]>a[r]&&a[r]>m)
        return false;
    else if(a[l]==a[r])
        return check(l+1,r-1,m);
    else
        return false;
}
char ch[maxn];
int main()
{
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&a[i]);
    }
    int l=1;
    int r=n;
    int len=0;
    int m=-1;
    while(l<=r)
    {
        //cout<<l<<" "<<r<<" "<<s<<endl;
        if(a[l]<=m&&a[r]<=m)
            break;
        if(a[l]>m&&a[r]<=m)
        {
            m=a[l];
            ch[++len]='L';
            l++;
        }
        else if(a[l]<=m&&a[r]>m)
        {
            m=a[r];
            ch[++len]='R';
            r--;
        }
        else if(a[l]>m&&a[r]>m)
        {
            if(a[l]<a[r])
            {
                m=a[l];
                ch[++len]='L';
                l++;
            }
            else if(a[r]<a[l])
            {
                m=a[r];
                ch[++len]='R';
                r--;
            }
            else
            {
                int mid=a[l];
                int l1=l+1;
                int r1=r-1;
                while(l1<r&&a[l1]>mid)
                {
                    mid=a[l1];
                    l1++;
                }
                mid=a[r];
                while(r1>l&&a[r1]>mid)
                {
                    mid=a[r1];
                    r1--;
                }
                if(l1-l>=r-r1)
                {
                    m=a[l];
                    ch[++len]='L';
                    l++;
                }
                else
                {
                    m=a[r];
                    ch[++len]='R';
                    r--;
                }
            }
        }
    }
    printf("%d\n",len);
    for(int i=1; i<=len; i++)
    {
        printf("%c",ch[i]);
    }
}