暴力解题的暴力题解……我以为会超时，结果它放过了我……

下面是题目描述：

There is a game called "Unique Bid Auction". You can read more about it here: https://en.wikipedia.org/wiki/Unique_bid_auction (though you don't have to do it to solve this problem).

Let's simplify this game a bit. Formally, there are n participants, the i-th participant chose the number ai. The winner of the game is such a participant that the number he chose is unique (i. e. nobody else chose this number except him) and is minimal (i. e. among all unique values of a the minimum one is the winning one).

Your task is to find the index of the participant who won the game (or -1 if there is no winner). Indexing is 1-based, i. e. the participants are numbered from 1 to n.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1≤t≤2⋅104) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1≤n≤2⋅105) — the number of participants. The second line of the test case contains n integers a1,a2,…,an (1≤ai≤n), where ai is the i-th participant chosen number.

It is guaranteed that the sum of n does not exceed 2⋅105 (∑n≤2⋅105).

Output

For each test case, print the answer — the index of the participant who won the game (or -1 if there is no winner). Note that the answer is always unique.

Example

input

1 1

2 1 3

2 2 2 3

2 3 2 4 2

1 1 5 5 4 4

output

-1

下面是解题过程：

从本题来看，需要选取有且仅有一的最小值。所以，我们首先需要判断最小值，然后判断是否唯一。
在输入过程中，我们需要统计该数字一共出现了几次。
将数组的序号用另外一个数组存入，然后将该数组排序，获得最小值。
判断该最小值是否唯一。
如果上述条件不满足，那么输出-1；如果满足，则找到在原数组对应的编号进行输出。

下面是AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>


const int N=1e5*3;
using namespace std;


int n;
int a[N],b[N];
int main()
{
    int t;
    scanf("%d",&t);
    while (t--)
    {
        int c[N]= {0};
        scanf("%d",&n);
        for(int i=1; i<n+1; i++)
        {
            scanf("%d",&a[i]);
            b[i]=a[i];
            c[a[i]]++;
        }
        sort(a+1,a+n+1);

        int num=-1,flag=0;
        for(int i=1; i<n+1; i++)
        {
            if(c[a[i]]==1)
            {
                num=a[i];
                flag=1;
                break;
            }
        }

        if(flag==0) printf("-1\n");
        else
        {
            for(int i=1; i<n+1; i++)
            {
                if(b[i]==num)
                {
                    printf("%d\n",i);
                    break;
                }
            }
        }

    }

    return 0;
}

【题解】Unique Bid Auction

下面是题目描述：

下面是解题过程：

下面是AC代码：