Codeforces Round #445 B. Vlad and Cafes【时间轴】

B. Vlad and Cafes

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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.

Output

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

Examples

input

5
1 3 2 1 2

output

3

input

6
2 1 2 2 4 1

output

2

Note

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+10;
const int inf = 9999999;
int vis[maxn];
int main()
{
    int n,a;
    int min;
    int m;
    while(cin>>n)
    {
        min=1000000,m=-1;
        for(int i=1;i<=n;i++)
        {
            cin>>a;
            vis[a]=i;
        }

        for(int i=0;i<=200000;i++)
        {
            if(vis[i]&&vis[i]<min)
            {
               min=vis[i];
               m=i;
            }
        }
        cout<<m<<endl;

    }
}