Problem: In 2N - 1 boxes there are apples and oranges. Your task is to choose N boxes so, that they will contain not less than half of all the apples and not less than half of all the oranges.

Input

The first input line contains one number T — amount of tests. The description of each test starts with a natural number N — amount of boxes. Each of the following 2N - 1 lines contains numbers ai and oi — amount of apples and oranges in the i-th box (0 ≤ ai, oi ≤ 109). The sum of N in all the tests in the input doesn't exceed 105. All the input numbers are integer.

Output

For each test output two lines. In the first line output YES, if it's possible to choose N boxes, or NO otherwise. If the answer is positive output in the second line N numbers — indexes of the chosen boxes. Boxes are numbered from 1 in the input order. Otherwise leave the second line empty. Separate the numbers with one space.

Examples

Input

2
2
10 15
5 7
20 18
1
0 0


Output

YES
1 3
YES
1

Solution (in PYTHON3):

Solution:
import os,io
from sys import stdout
import collections
# import random
import math
from operator import itemgetter
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from collections import Counter
# from decimal import Decimal
# import heapq
# from functools import lru_cache

# import sys
# sys.setrecursionlimit(10**6)

# from functools import lru_cache
# @lru_cache(maxsize=None)

def primes(n):
    sieve = [True] * n
    for i in range(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
    return [2] + [i for i in range(3,n,2) if sieve[i]]

def binomial_coefficient(n, k):
    if 0 <= k <= n:
        ntok = 1
        ktok = 1
        for t in range(1, min(k, n - k) + 1):
            ntok *= n
            ktok *= t
            n -= 1
        return ntok // ktok
    else:
        return 0

def powerOfK(k, max):
    if k == 1:
        return [1]
    if k == -1:
        return [-1, 1]

    result = []
    n = 1
    while n <= max:
        result.append(n)
        n *= k
    return result

def prefixSum(arr):
    for i in range(1, len(arr)):
        arr[i] = arr[i] + arr[i-1]
    return arr

def divisors(n):
    i = 1
    result = []
    while i*i <= n:
        if n%i == 0:
            if n/i == i:
                result.append(i)
            else:
                result.append(i)
                result.append(n/i)
        i+=1
    return result

def kadane(a,size):
    max_so_far = 0
    max_ending_here = 0

    for i in range(0, size):
        max_ending_here = max_ending_here + a[i]
        if (max_so_far < max_ending_here):
            max_so_far = max_ending_here

        if max_ending_here < 0:
            max_ending_here = 0
    return max_so_far

# @lru_cache(maxsize=None)
def digitsSum(n):
    if n == 0:
        return 0
    r = 0
    while n > 0:
        r += n % 10
        n //= 10
    return r

def print_grid(grid):
    for line in grid:
        print("".join(line))

# INPUTS  --------------------------
# s = input().decode('utf-8').strip()
# n = int(input())
# l = list(map(int, input().split()))
# t = int(input())
# for _ in range(t):
# for _ in range(t):

t = int(input())
for _ in range(t):
    n = int(input())
    boxes = []
    i = 0
    for _ in range(2*n-1):
        boxes.append(list(map(int, input().split()))+[i+1])
        i+=1
    boxes = sorted(boxes, key=itemgetter(1))

    even = boxes[::2]
    odd = boxes[1::2] + [boxes[-1]]

    apples = sum([e[0] for e in boxes])
    halfapple = apples // 2
    if apples % 2 == 1:
        halfapple += 1

    aeven = sum([e[0] for e in even])
    aodd = sum([e[0] for e in odd])

    if aeven >= halfapple:
        print('YES')
        res = []
        for e in even:
            res.append(e[2])
        print(" ".join([str(e) for e in sorted(res)]))
    elif aodd >= halfapple:
        print('YES')
        res = []
        for e in odd:
            res.append(e[2])

        print(" ".join([str(e) for e in sorted(res)]))
    else:
        print('NO')