#include<stdio.h>
#include<iostream>
#include<string.h>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<algorithm>
#include<stdlib.h>
#include<string>
#include<stack>
#include<set>
using namespace std;
#define pb push_back
#define POP pop_back()
#define DEBUG cout << "DEBUG\n";
#define ll long long
#define db double
#define endl '\n'
#define fo(i,a,b) for(int i = a;i <= b;i++)
#define lowbit(x) (x&-x);
const int maxn = 5e5+5;
const int inf = 0x3f3f3f3f;
const ll inF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1e8;
const db eps = 1e-8;
struct node{
	int v , id;
	bool operator < (const node &b)const{
		return v < b.v;
	}
};
node cnt[maxn];
int a[maxn];
int main(){
	int t , n;cin >> t;
	while(t--){
		cin >> n;
		for(int i = 1;i <= n;i++)cnt[i].v = cnt[i].id = 0;
		for(int i = 1;i <= n;i++)scanf("%d",&a[i]) , cnt[a[i]].v++ , cnt[a[i]].id = i;
		int ans = -1;
		for(int i = 1;i <= n;i++){
			if(cnt[i].v == 1){ans = cnt[i].id;break;}
		}
		if(ans == -1){
			cout << "-1\n";
		}
		else cout << ans << endl;
	}
	return 0;
}
### Prompt

In Cpp, your task is to solve the following problem:

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 a_i. 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 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of participants. The second line of the test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n), where a_i is the i-th participant chosen number.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

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


6
2
1 1
3
2 1 3
4
2 2 2 3
1
1
5
2 3 2 4 2
6
1 1 5 5 4 4


Output


-1
2
4
1
2
-1

### Response

```cpp
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<algorithm>
#include<stdlib.h>
#include<string>
#include<stack>
#include<set>
using namespace std;
#define pb push_back
#define POP pop_back()
#define DEBUG cout << "DEBUG\n";
#define ll long long
#define db double
#define endl '\n'
#define fo(i,a,b) for(int i = a;i <= b;i++)
#define lowbit(x) (x&-x);
const int maxn = 5e5+5;
const int inf = 0x3f3f3f3f;
const ll inF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1e8;
const db eps = 1e-8;
struct node{
	int v , id;
	bool operator < (const node &b)const{
		return v < b.v;
	}
};
node cnt[maxn];
int a[maxn];
int main(){
	int t , n;cin >> t;
	while(t--){
		cin >> n;
		for(int i = 1;i <= n;i++)cnt[i].v = cnt[i].id = 0;
		for(int i = 1;i <= n;i++)scanf("%d",&a[i]) , cnt[a[i]].v++ , cnt[a[i]].id = i;
		int ans = -1;
		for(int i = 1;i <= n;i++){
			if(cnt[i].v == 1){ans = cnt[i].id;break;}
		}
		if(ans == -1){
			cout << "-1\n";
		}
		else cout << ans << endl;
	}
	return 0;
}
```