Vlad likes to eat in cafes very much. During his life, he has visited cafes n times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.

First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe.

In first line there is one integer n (1 ≤ n ≤ 2·105) — number of cafes indices written by Vlad.

In second line, n numbers a1, a2, ..., an (0 ≤ ai ≤ 2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted.

Print one integer — index of the cafe that Vlad hasn't visited for as long as possible.

In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.

In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes.

【题意】： 找到一个数满足，它最后一次出现的位置是i，位置i+1到n所有其他数字都出现过至少一次 。

【分析】： 1.上一次当彼佳参观每一家咖啡馆时，将其放在数组中。 2.现在你需要找到这个数组中最小的位置并打印它。

【代码】：

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1e5+1e5+;

const int inf = ;

int vis[maxn];

int main()

{

    int n,a;

    int min;

    int m;

    while(cin>>n)

    {

        min=,m=-;

        for(int i=;i<=n;i++)

        {

            cin>>a;

            vis[a]=i;

        }

        for(int i=;i<=;i++)

        {

            if(vis[i]&&vis[i]<min)

            {

               min=vis[i];

               m=i;

            }

        }

        cout<<m<<endl;

    }

}