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:
# Author Name: Ajay Meena
# Codeforce : https://codeforces.com/profile/majay1638

# -------- IMPORTANT ---------#
# SUN BHOS**KE AGAR MERA TEMPLATE COPY KAR RHA HAI NA TOH KUCH CHANGES BHI KAR DENA ESS ME, VARNA MUJEHY WARNING AAYEGI BAAD ME, PLEASE YAAR KAR DENA, OK :).
import sys
import math
from sys import stdin, stdout

# //Most Frequently Used Number Theory Concepts
# VAISE MEIN JAYDA USE KARTA NHI HU ENHE BUT COOL BANNE KE LIYE LIKH LEYA TEMPLATE ME VARNA ME YE TOH DUSRI FILE MAI SE BHI COPY PASTE KAR SAKTA THA :).


def sieve(N):
    primeNumbers = [True]*(N+1)
    primeNumbers[0] = False
    primeNumbers[1] = False
    i = 2
    while i*i <= N:
        j = i
        if primeNumbers[j]:
            while j*i <= N:
                primeNumbers[j*i] = False
                j += 1
        i += 1
    return primeNumbers


def getPrime(N):
    primes = sieve(N)
    result = []
    for i in range(len(primes)):
        if primes[i]:
            result.append(i)
    return result


def factor(N):
    factors = []
    i = 1
    while i*i <= N:
        if N % i == 0:
            factors.append(i)
            if i != N//i:
                factors.append(N//i)
        i += 1
    return sorted(factors)


def gcd(a, b):
    if a < b:
        return gcd(b, a)
    if b == 0:
        return a
    return gcd(b, a % b)


def extendedGcd(a, b):
    if a < b:
        return extendedGcd(b, a)
    if b == 0:
        return [a, 1, 0]
    res = extendedGcd(b, a % b)
    x = res[2]
    y = res[1]-(math.floor(a/b)*res[2])
    res[1] = x
    res[2] = y
    return res


def iterativeModularFunc(a, b, c):
    res = 1
    while b > 0:
        if b & 1:
            res = (res*a) % c
        a = (a*a) % c
        b = b//2
    return res

# TAKE INPUT
# HAAN YE BHUT KAAM AATA HAI INPUT LENE ME


def get_ints_in_variables():
    return map(int, sys.stdin.readline().strip().split())


def get_int(): return int(input())


def get_ints_in_list(): return list(
    map(int, sys.stdin.readline().strip().split()))


def get_list_of_list(n): return [list(
    map(int, sys.stdin.readline().strip().split())) for _ in range(n)]


def get_string(): return sys.stdin.readline().strip()


def canWin(arr, prefix, n, mid):
    s = prefix[mid+1]
    for i in range(mid+1, n):
        if s >= arr[i][0]:
            s += arr[i][0]
        else:
            break
    if s >= prefix[-1]:
        return True
    else:
        return False


def Solution(arr, n):
    l = 0
    r = n-1
    a = sorted([[arr[i], i+1] for i in range(n)], key=lambda x: x[0])
    prefix = [0]
    for v in a:
        prefix.append(prefix[-1]+v[0])

    ans = -1
    # print(prefix)
    while l <= r:
        mid = (l+r)//2

        if canWin(a, prefix, n, mid):
            ans = mid
            r = mid-1
        else:
            l = mid+1
    if ans == -1:

        print(n)
        for v in range(1, n+1):
            print(v, end=" ")
        print()
        return
    else:
        print(n-(ans))
        res = []
        for v in range(ans, n):
            res.append(a[v][1])
        res = sorted(res)
        for v in res:
            print(v, end=" ")
        print()


def main():
    # //Write Your Code Here
    for _ in range(get_int()):
        n = get_int()
        arr = get_ints_in_list()
        Solution(arr, n)


#  calling main Function
if __name__ == "__main__":
    main()