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 JAVA):

Solution:
//never give up
//Creator: arunm 16-02-2021 19:31

import java.util.*;
import java.io.*;

public class TaskE {
    FastReader fr = new FastReader();
    StringBuilder sb = new StringBuilder();
    int mod = (int) (1e9 + 7);
    int inf = (int) (1e7 + 6);

    //code here
    public void solver() {
       int tt=fr.nextInt();
       while(tt-->0){
           int n = fr.nextInt();
           int a[]=new int[n];
           for(int i=0;i<n;i++){
               a[i]=fr.nextInt();
           }
           int [][] ai = new int[n][];
           for(int i=0;i<n;i++){
               ai[i]=new int[]{a[i],i};
           }
           Arrays.sort(ai,(x,y)->{
              if(x[0]!=y[0]) return x[0]-y[0];
              return x[1]-y[1];
           });
           long cum[]=new long[n+1];
           for(int i=0;i<n;i++){
               cum[i+1]=cum[i]+ai[i][0];
           }
           int from=0;
           for(int i=n-2;i>=0;i--){
               if(cum[i+1]<ai[i+1][0]){
                   from = i+1;
                   break;
               }
           }
           boolean []res = new boolean[n];
           for(int i=from;i<n;i++){
               res[ai[i][1]]=true;
           }
           sb.append(n-from+"\n");
           for(int i=0;i<n;i++){
               if(res[i]){
                   sb.append((i+1)+" ");
               }
           }
           sb.append("\n");
       }
        System.out.println(sb);
    }
    public static void main(String[] args) {
        new TaskE().solver();
    }

    /* Helper Methods begin*/

    //copied from SecondThread
    public String sortString(String s) {
        char ch[] = s.toCharArray();
        Arrays.sort(ch);
        return String.valueOf(ch);
    }

    public void ruffleSort(int[] a) {
        //ruffle
        int n = a.length;
        Random r = new Random();
        for (int i = 0; i < a.length; i++) {
            int oi = r.nextInt(n), temp = a[i];
            a[i] = a[oi];
            a[oi] = temp;
        }
        //then sort
        Arrays.sort(a);
    }

    public long power(long a, long b) {
        if (a == 1 || b == 0) {
            return 1;
        }
        long res = 1;
        while (b != 0) {
            if (b % 2 == 1) {
                res = (res * a);
            }
            b /= 2;
            a = (a * a);
        }
        return res;
    }

    public long modPower(long a, long b, long mod) {
        if (a == 1 || b == 0) {
            return 1;
        }
        long res = 1;
        while (b != 0) {
            if (b % 2 == 1) {
                res = (res * a) % mod;
            }
            b /= 2;
            a = (a * a) % mod;
        }
        return res % mod;
    }

    public long modInverse(long a, long mod) {
        return modPower(a, mod - 2, mod);
    }

    public long ncr(int n, int r) {
        long p = 1, k = 1;

        // C(n, r) == C(n, n-r),
        // choosing the smaller value
        if (n - r < r) {
            r = n - r;
        }

        if (r != 0) {
            while (r > 0) {
                p *= n;
                k *= r;

                // gcd of p, k
                long m = gcd(p, k);

                // dividing by gcd, to simplify
                // product division by their gcd
                // saves from the overflow
                p /= m;
                k /= m;

                n--;
                r--;
            }

            // k should be simplified to 1
            // as C(n, r) is a natural number
            // (denominator should be 1 ) .
        } else {
            p = 1;
        }

        // if our approach is correct p = ans and k =1
        return p;
    }

    public int lowerBound(int a[], int key) {
        int low = 0, high = a.length - 1;
        int ans = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (a[mid] >= key) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }

    public int upperBound(int a[], int key) {
        int low = 0, high = a.length - 1;
        int ans = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (a[mid] >= key) {
                ans = mid;
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return ans;
    }

    public int gcd(int a, int b) {
        if (a == 0) {
            return b;
        } else {
            return gcd(b % a, a);
        }
    }

    public long gcd(long a, long b) {
        if (a == 0) {
            return b;
        } else {
            return gcd(b % a, a);
        }
    }

    public int lcm(int a, int b) {
        long res = a * b;
        res /= gcd(a, b);
        return (int) res;
    }

    public long lcm(long a, long b) {
        long res = a * b;
        res /= gcd(a, b);
        return res;
    }

    public int max(int... a) {
        int res = Integer.MIN_VALUE;
        for (int x : a) {
            res = Math.max(res, x);
        }
        return res;
    }

    public long max(long... a) {
        long res = Long.MIN_VALUE;
        for (long x : a) {
            res = Math.max(res, x);
        }
        return res;
    }

    public int min(int... a) {
        int res = Integer.MAX_VALUE;
        for (int x : a) {
            res = Math.min(res, x);
        }
        return res;
    }

    public long min(long... a) {
        long res = Long.MAX_VALUE;
        for (long x : a) {
            res = Math.min(res, x);
        }
        return res;
    }

    /*Helper method ends*/

    /* FAST IO */
    class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    System.out.println(e);
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public String nextLing() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                System.out.println(e);
            }
            return str;
        }
    }

    //https://github.com/EgorKulikov/yaal/blob/master/lib/main/net/egork/collections/Pair.java
    class Pair<U, V> implements Comparable<Pair<U, V>> {
        public final U first;
        public final V second;

        public <U, V> Pair<U, V> makePair(U first, V second) {
            return new Pair<U, V>(first, second);
        }

        private Pair(U first, V second) {
            this.first = first;
            this.second = second;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) {
                return true;
            }
            if (o == null || getClass() != o.getClass()) {
                return false;
            }

            Pair pair = (Pair) o;

            return !(first != null ? !first.equals(pair.first) : pair.first != null) &&
                    !(second != null ? !second.equals(pair.second) : pair.second != null);
        }

        @Override
        public int hashCode() {
            int result = first != null ? first.hashCode() : 0;
            result = 31 * result + (second != null ? second.hashCode() : 0);
            return result;
        }

        public Pair<V, U> swap() {
            return makePair(second, first);
        }

        @Override
        public String toString() {
            return "(" + first + "," + second + ")";
        }

        @SuppressWarnings({"unchecked"})
        public int compareTo(Pair<U, V> o) {
            int value = ((Comparable<U>) first).compareTo(o.first);
            if (value != 0) {
                return value;
            }
            return ((Comparable<V>) second).compareTo(o.second);
        }
    }
}