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 PYTHON3):

Solution:
from collections import defaultdict,OrderedDict,Counter
from sys import stdin,stdout
from bisect import bisect_left,bisect_right
# import numpy as np
from queue import Queue,PriorityQueue
from heapq import heapify,heappop,heappush
from statistics import median,mean
from math import gcd,sqrt,floor,factorial,ceil,log2,log10,log
import fractions
import copy
from copy import deepcopy
import sys
import io
sys.setrecursionlimit(10**8)
import math
import os
import bisect
import collections
mod=pow(10,9)+7
import random
from random import random,randint,randrange
from time import  time;
def ncr(n, r, p=mod):
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den,
            p - 2, p)) % p
def normalncr(n,r):
    r=min(r,n-r)
    count=1;
    for i in range(n-r,n+1):
        count*=i;

    for i in range(1,r+1):
        count//=i;
    return count

inf=float("inf")
adj=defaultdict(set)
visited=defaultdict(int)
def addedge(a,b):
    adj[a].add(b)
    adj[b].add(a)
def bfs(v):
    q=Queue()
    q.put(v)
    visited[v]=1
    while q.qsize()>0:
        s=q.get_nowait()
        print(s)
        for i in adj[s]:
            if visited[i]==0:
                q.put(i)
                visited[i]=1
def dfs(v,visited):
    if visited[v]==1:
        return;
    visited[v]=1

    print(v)
    for i in adj[v]:
        dfs(i,visited)
# a9=pow(10,6)+10
# prime = [True for i in range(a9 + 1)]
# def SieveOfEratosthenes(n):
#     p = 2
#     while (p * p <= n):
#         if (prime[p] == True):
#             for i in range(p * p, n + 1, p):
#                 prime[i] = False
#         p += 1
# SieveOfEratosthenes(a9)
# prime_number=[]
# for i in range(2,a9):
#     if prime[i]:
#         prime_number.append(i)
def reverse_bisect_right(a, x, lo=0, hi=None):
    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if x > a[mid]:
            hi = mid
        else:
            lo = mid+1
    return lo
def reverse_bisect_left(a, x, lo=0, hi=None):
    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if x >= a[mid]:
            hi = mid
        else:
            lo = mid+1
    return lo

def get_list():
    return list(map(int,input().split()))
def get_str_list_in_int():
    return [int(i) for i in list(input())]
def get_str_list():
    return list(input())
def get_map():
    return map(int,input().split())
def input_int():
    return int(input())
def matrix(a,b):
    return [[0 for i in range(b)] for j in range(a)]
def swap(a,b):
    return b,a
def find_gcd(l):
    a=l[0]
    for i in range(len(l)):
        a=gcd(a,l[i])
    return a;
def is_prime(n):
    sqrta=int(sqrt(n))
    for i in range(2,sqrta+1):
        if n%i==0:
            return 0;
    return 1;
def prime_factors(n):
    sqrta = int(sqrt(n))
    for i in range(2,sqrta+1):
        if n%i==0:
            return [i]+prime_factors(n//i)
    return [n]
def p(a):
    if type(a)==str:
        print(a+"\n")
    else:
        print(str(a)+"\n")
def ps(a):
    if type(a)==str:
        print(a)
    else:
        print(str(a))
def kth_no_not_div_by_n(n,k):
    return k+(k-1)//(n-1)
nc="NO"
yc="YES"
ns="No"
ys="Yes"

# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# input=stdin.readline
# print=stdout.write
def get_value(m,index):
    if 0<=index<len(m):
        return m[index]
    return 0;

t=int(input())
for i in range(t):
    n=int(input())
    l=get_list();
    suma = sum(l)
    l=[[l[i],i+1] for i in range(n)]

    l.sort()
    dict=defaultdict(int)
    dict[l[-1][1]]=1

    for i in reversed(range(n)):
        suma-=l[i][0]

        if suma>=l[i][0]:
            dict[l[i-1][1]]=1;
        else:
            break;
    gamma=sorted(list(dict))
    print(len(gamma))
    print(*gamma)
# 6
# 1 1 1 9 16 18
# 1 2 5 6