Problem Description

AFA want to distribution her money to somebody.She divide her money into n same parts.One who want to get the money can get more than one part.But if one man's money is more than the sum of all others'.He shoule be punished.Each one who get a part of money would write down his ID on that part.

Input

There are multiply cases.For each case,there is a single integer n(1<=n<=1000) in first line.In second line,there are n integer a1,a2...an(0<=ai<10000)ai is the the ith man's ID.

Output

Output ID of the man who should be punished.If nobody should be punished,output -1.

Sample Input

3
1 1 2
4
2 1 4 3

Sample Output

1
-1

思路：本题数据量不大，也不需要排序什么的，直接装上数组比较就ac

#include <stdio.h>
#include <string.h>
#define maxn 10005
int ans[maxn];
int main(){
int n;
while(scanf("%d",&n)!=EOF){
memset(ans,0,sizeof(ans));
for(int i=0;i<n;i++){
int temp;
scanf("%d",&temp);
ans[temp]++;
}
int ok=0;
for(int i=0;i<maxn;i++){
if(ans[i]&&ans[i]>(n-ans[i])){
printf("%d\n",i);
ok=1;
break;
}
}
if(!ok)
printf("-1\n");
}
return 0;
}