/*
	Solution by:- 
	--------------Sannidhay Vashal
	----------------NIT SRINAGAR
 
*/
#include<iostream>
#include<vector>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<math.h>
#include<climits>
#include<set>
#include<cstring>
#include<unordered_map>
#include <cstdlib> 
#include<cmath>
 
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL)
#define ll            long long int
#define ull           unsigned long long int
#define cinll(x)      ll x;cin>>x;
#define cini(x)       int x;cin>>x;
#define cins(x)       string x;cin>>x;
#define vect(x)       vector<ll> x;
#define vect1(x)      vector<ll> x;x.push_back(0);
#define pb(x)         push_back(x)
#define mp(x,y)       make_pair(x,y)
#define MAX           1e18
#define MIN           -1000000007
#define MOD           1000000007

using namespace std;

bool compare(pair<ll,ll> a,pair<ll,ll> b){
  return a.second<b.second;
}

int main(){
  cinll(t);
  while(t--){
    cinll(n);
    vector<ll> arr;
    unordered_map<ll,ll> mapit;
    vector<pair<ll,ll>> ans;
    for(ll i=0;i<n;i++) {
      cinll(x);
      arr.pb(x);
      mapit[x]++;
    }
    ll anss = -1;
    for(ll i=0;i<n;i++){
      if(mapit[arr[i]]==1) ans.pb(mp(i+1,arr[i]));
    }
    sort(ans.begin(),ans.end(),compare);
    if(ans.size()>=1) anss = ans[0].first;
    cout<<anss<<"\n";
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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
/*
	Solution by:- 
	--------------Sannidhay Vashal
	----------------NIT SRINAGAR
 
*/
#include<iostream>
#include<vector>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<math.h>
#include<climits>
#include<set>
#include<cstring>
#include<unordered_map>
#include <cstdlib> 
#include<cmath>
 
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL)
#define ll            long long int
#define ull           unsigned long long int
#define cinll(x)      ll x;cin>>x;
#define cini(x)       int x;cin>>x;
#define cins(x)       string x;cin>>x;
#define vect(x)       vector<ll> x;
#define vect1(x)      vector<ll> x;x.push_back(0);
#define pb(x)         push_back(x)
#define mp(x,y)       make_pair(x,y)
#define MAX           1e18
#define MIN           -1000000007
#define MOD           1000000007

using namespace std;

bool compare(pair<ll,ll> a,pair<ll,ll> b){
  return a.second<b.second;
}

int main(){
  cinll(t);
  while(t--){
    cinll(n);
    vector<ll> arr;
    unordered_map<ll,ll> mapit;
    vector<pair<ll,ll>> ans;
    for(ll i=0;i<n;i++) {
      cinll(x);
      arr.pb(x);
      mapit[x]++;
    }
    ll anss = -1;
    for(ll i=0;i<n;i++){
      if(mapit[arr[i]]==1) ans.pb(mp(i+1,arr[i]));
    }
    sort(ans.begin(),ans.end(),compare);
    if(ans.size()>=1) anss = ans[0].first;
    cout<<anss<<"\n";
  }
  return 0;
}

```