Petya has an array a consisting of n integers. He wants to remove duplicate (equal) elements.
Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.
Input
The first line contains a single integer n (1≤n≤50) — the number of elements in Petya’s array.
The following line contains a sequence a1,a2,…,an (1≤ai≤1000) — the Petya’s array.
Output
In the first line print integer x — the number of elements which will be left in Petya’s array after he removed the duplicates.
In the second line print x integers separated with a space — Petya’s array after he removed the duplicates. For each unique element only the rightmost entry should be left.
Examples
Input
6
1 5 5 1 6 1
Output
3
5 6 1
Input
5
2 4 2 4 4
Output
2
2 4
Input
5
6 6 6 6 6
Output
1
6
Note
In the first example you should remove two integers 1, which are in the positions 1 and 4. Also you should remove the integer 5, which is in the position 2.
In the second example you should remove integer 2, which is in the position 1, and two integers 4, which are in the positions 2 and 4.
In the thir example you should remove four integers 6, which are in the positions 1, 2, 3 and 4.

删除重复的，保留最右边的，所以从右边开始遍历

代码：

#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
int a[55];
bool vis[1005];
vector<int> res;
int main(void){
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    for(int i=n-1;i>=0;i--){
        if(!vis[a[i]]){
            res.push_back(a[i]);
            vis[a[i]]=true;
        }
    }
    int l=res.size();
    printf("%d\n",l);
    for(int i=l-1;i>0;i--){
        printf("%d ",res[i]);
    }
    printf("%d\n",res[0]);
}