Problem: A championship is held in Berland, in which n players participate. The player with the number i has a_i (a_i ≥ 1) tokens.

The championship consists of n-1 games, which are played according to the following rules:

  * in each game, two random players with non-zero tokens are selected; 
  * the player with more tokens is considered the winner of the game (in case of a tie, the winner is chosen randomly); 
  * the winning player takes all of the loser's tokens; 



The last player with non-zero tokens is the winner of the championship.

All random decisions that are made during the championship are made equally probable and independently.

For example, if n=4, a = [1, 2, 4, 3], then one of the options for the game (there could be other options) is: 

  * during the first game, the first and fourth players were selected. The fourth player has more tokens, so he takes the first player's tokens. Now a = [0, 2, 4, 4]; 
  * during the second game, the fourth and third players were selected. They have the same number of tokens, but in a random way, the third player is the winner. Now a = [0, 2, 8, 0]; 
  * during the third game, the second and third players were selected. The third player has more tokens, so he takes the second player's tokens. Now a = [0, 0, 10, 0]; 
  * the third player is declared the winner of the championship. 



Championship winners will receive personalized prizes. Therefore, the judges want to know in advance which players have a chance of winning, i.e have a non-zero probability of winning the championship. You have been asked to find all such players. 

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The first line of each test case consists of one positive integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of players in the championship.

The second line of each test case contains n positive integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the number of tokens the players have.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5. 

Output

For each test case, print the number of players who have a nonzero probability of winning the championship. On the next line print the numbers of these players in increasing order. Players are numbered starting from one in the order in which they appear in the input. 

Example

Input


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


Output


3
2 3 4 
5
1 2 3 4 5 

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <cmath>
#include <list>
#include <map>
#include <algorithm>
#include <climits>
#include <sstream> 
#include <iomanip>
#include <bits/stdc++.h>
#include <cstdio>
#include <bitset>
#include <cstring>
#include <deque>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <string>
using namespace std;
#define ll long long int
#define cbit __builtin_popcountll
#define pl pair<long, long>
#define pqp priority_queue<pl>
// getline(cin, str); 
//  __gcd(a,b)
// vec.insert(vec.begin()+i,value); log2(n); tolower(s[i]);
// ostringstream s; s<<i; string ans =s.str();
// string c=to_string(i);   s.push_back(c);
// next_permutation(nums.begin(),nums.end())
// priority_queue<PII, vector<PII>, greater<PII>> pq;
/* Iterative Function to calculate (x^y)%p
in O(log y) */
unsigned long long power(unsigned long long x, int y, int p)
{
    unsigned long long res = 1; // Initialize result
 
    x = x % p; // Update x if it is more than or
    // equal to p
 
    while (y > 0)
    {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res * x) % p;
 
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
bool cmp(pair<string, int> &a, pair<string, int> &b)
{
    return a.second < b.second;
}
void sort(map<string, int> &M)
{
 
    // Declare vector of pairs
    vector<pair<string, int> > A;
 
    // Copy key-value pair from Map
    // to vector of pairs
    for (auto &it : M)
    {
        A.push_back(it);
    }
 
    // Sort using comparator function
    sort(A.begin(), A.end(), cmp);
 
    // Print the sorted value
    for (auto &it : A)
    {
 
        cout << it.first << ' '
             << it.second << endl;
    }
}
// Returns n^(-1) mod p
unsigned long long modInverse(unsigned long long n, int p)
{
    return power(n, p - 2, p);
}
 
// Returns nCr % p using Fermat's little
// theorem.
unsigned long long nCrModPFermat(unsigned long long n, int r, int p)
{
    // If n<r, then nCr should return 0
    if (n < r)
        return 0;
    // Base case
    if (r == 0)
        return 1;
 
    // Fill factorial array so that we
    // can find all factorial of r, n
    // and n-r
    unsigned long long fac[n + 1];
    fac[0] = 1;
    for (int i = 1; i <= n; i++)
        fac[i] = (fac[i - 1] * i) % p;
 
    return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) % p;
}
long long gcd(long long int a, long long int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
long long lcm(ll a, ll b)
{
    return (a / gcd(a, b)) * b;
}
 
bool isPrime(ll n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (ll i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
ll find(ll s, ll n, ll pos)
{
    ll mid = s + n / 2;
    if (mid == pos)
    {
        return 1;
    }
    else if (mid > pos)
    {
        return 1 + find(mid + 1, n, pos);
    }
    else
    {
        return 1 + find(s, mid, pos);
    }
}
// ll nCr(ll n, ll r) 
// { 
//     return fact(n) / (fact(r) * fact(n - r)); 
// } 
ll fact(ll n,ll p)
{
    // single line to find factorial
    return (n == 1 || n == 0) ? 1 : (n * (fact(n - 1,p)));
}
ll num(ll n, ll ans)
{
    n /= 10;
    if (n == 0)
    {
        return ans;
    }
    else
    {
        return ans + num(n, ans + 1);
    }
}
ll nof(ll n)
{
    if (n == 1)
    {
        return 1;
    }
    ll ans = 2;
    for (ll i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
        {
            if (i * i != n)
            {
                ans += 2;
            }
            else
            {
                ans++;
            }
        }
    }
    return ans;
}
ll getsum(ll n)
{
    ll ans = 0;
    while (n)
    {
        ans += n % 10;
        n /= 10;
    }
    return ans;
}
ll power(ll num, ll deg,ll m) {
    if (!deg)
        return 1;
 
    if (deg % 2) {
        return (power(num, deg - 1,m) * num) % m;     
    }
    else {
        ll sqrt_res = power(num, deg / 2,m);
        return (sqrt_res * sqrt_res) % m;
    }
}
// vector<ll> prime;
void prm(bool mark[100100],ll size)
{
    mark[0]=true;
    mark[1]=true;
    for(ll i=2;i*i<size;i++)
    {
        if(mark[i]==false)
        {
            // prm.push_back(i);
            for(ll j=i+i;j<size;j+=i)
            {
                mark[j]=true;
            }
        }
    }
}
void finddp(vector<ll>& ans,vector<ll> vec,ll d,ll s,ll e )
{
    if(s>e)
    {
        return;
    }
    ll min=-1000000000000;
    ll mi=s;
    for(ll i=s;i<=e;i++)
    {
        if(vec[i]>min)
        {
            mi=i;
            min=vec[i];
        }
    }
    ans[mi]=d;
    // cout<<mi<<"\n";
    finddp(ans,vec,d+1,s,mi-1);
    finddp(ans,vec,d+1,mi+1,e);
    return;
}

bool stt(const pair<ll,ll> &a, const pair<ll,ll> &b)
{
    return (a.first<b.first);
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    // #ifndef ONLINE_JUDGE
    // freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout);
    // #endif
    
    ll t = 1;
    cin >> t;
    while (t--)
    {
        ll n;
        cin>>n;
        
        vector<pair<ll,ll> > vec;
        for(ll i=0;i<n;i++)
        {
            ll p;
            cin>>p;
            vec.push_back(make_pair(p,i));
        }
        if(n==1)
        {
            cout<<1<<"\n";
            cout<<1<<"\n";
            continue;
        }
        sort(vec.begin(),vec.end(),stt);
        // for( auto itr:vec)
        // {
        //     cout<<itr.first<<" " ;
        // }
        // cout<<"\n";
        ll ans=1;
        ll temp=vec[0].first;
        for(ll i=1;i<n-1;i++)
        {
            if((vec[i].first+temp)<vec[i+1].first)
            {
                ans=i+1;
            }
            temp+=vec[i].first;
        }
        vector<ll> op;
        if(vec[ans-1].first==vec[ans].first)
        {
            ans--;
        }
        for(ll i=ans;i<n;i++)
        {
            op.push_back(vec[i].second);
        }
        sort(op.begin(),op.end());
        cout<<op.size()<<"\n";
        for(auto itr :op)
        {
            cout<<itr+1<<" ";
        }
        cout<<'\n';
    }
 
    return 0;
}