You got a job as a marketer in a pet shop, and your current task is to boost sales of cat food. One of the strategies is to sell cans of food in packs with discounts. 

Suppose you decided to sell packs with a cans in a pack with a discount and some customer wants to buy x cans of cat food. Then he follows a greedy strategy: 

  * he buys \left⌊ x/a \right⌋ packs with a discount; 
  * then he wants to buy the remaining (x mod a) cans one by one. 



\left⌊ x/a \right⌋ is x divided by a rounded down, x mod a is the remainer of x divided by a.

But customers are greedy in general, so if the customer wants to buy (x mod a) cans one by one and it happens that (x mod a) ≥ a/2 he decides to buy the whole pack of a cans (instead of buying (x mod a) cans). It makes you, as a marketer, happy since the customer bought more than he wanted initially.

You know that each of the customers that come to your shop can buy any number of cans from l to r inclusive. Can you choose such size of pack a that each customer buys more cans than they wanted initially?

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first and only line of each test case contains two integers l and r (1 ≤ l ≤ r ≤ 10^9) — the range of the number of cans customers can buy.

Output

For each test case, print YES if you can choose such size of pack a that each customer buys more cans than they wanted initially. Otherwise, print NO.

You can print each character in any case.

Example

Input


3
3 4
1 2
120 150


Output


YES
NO
YES

Note

In the first test case, you can take, for example, a = 5 as the size of the pack. Then if a customer wants to buy 3 cans, he'll buy 5 instead (3 mod 5 = 3, 5/2 = 2.5). The one who wants 4 cans will also buy 5 cans.

In the second test case, there is no way to choose a.

In the third test case, you can take, for example, a = 80.
##############--->>>>> Deepcoder Amit Kumar Bhuyan <<<<<---##############

"""
 Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
from __future__ import division, print_function
 
import os,sys
from io import BytesIO, IOBase
 
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
 
 
def ii():  return int(input())
def si():  return input()
def mi():  return map(int,input().strip().split(" "))
def msi(): return map(str,input().strip().split(" "))
def li():  return list(mi())
 
def dmain():
    sys.setrecursionlimit(1000000)
    threading.stack_size(1024000)
    thread = threading.Thread(target=main)
    thread.start()
    
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import log,sqrt,factorial,cos,tan,sin,radians
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *
#import threading
#from itertools import permutations
#Copy 2D list  m = [x[:] for x in mark] .. Avoid Using Deepcopy
import sys
input = sys.stdin.readline
scanner = lambda: int(input())
string = lambda: input().rstrip()
get_list = lambda: list(read())
read = lambda: map(int, input().split())
get_float = lambda: map(float, input().split())
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;

 
def ceil(x):
    if x != int(x):
        x = int(x) + 1
    return x
 
def factorial(x, m):
	val = 1
	while x>0:
		val = (val * x) % m
		x -= 1
	return val

def fact(x):
	val = 1
	while x > 0:
		val *= x
		x -= 1
	return val
    
# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp;
 
## gcd function
def gcd(a,b):
    if b == 0:
        return a;
    return gcd(b, a % b);

## lcm function
def lcm(a, b):
	return (a * b) // math.gcd(a, b)

def is_integer(n):
	return math.ceil(n) == math.floor(n)
 
## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
	if k > n:
		return 0
	if(k > n - k):
		k = n - k
	res = 1
	for i in range(k):
		res = res * (n - i)
		res = res / (i + 1)
	return int(res) % MOD

## alternative
def comb(n,x):
    a,b=1,1
    for i in range(1,x+1):
        a*=(n+1-i)
        b*=i
    return (a//b)%MOD
 
## upper bound function code -- such that e in a[:i] e < x;

 
## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0 and n > 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0 and n > 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes
 
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
    if y == 0:
        return 1
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 
 
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b;
 
# find function with path compression included (recursive)
# def find(x, link):
#     if link[x] == x:
#         return x
#     link[x] = find(link[x], link);
#     return link[x];
 
# find function with path compression (ITERATIVE)
def find(x, link):
    p = x;
    while( p != link[p]):
        p = link[p];
    
    while( x != p):
        nex = link[x];
        link[x] = p;
        x = nex;
    return p;
 
 
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x
 
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    prime[0], prime[1] = False, False
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime

# Euler's Toitent Function phi
def phi(n) : 
  
    result = n 
    p = 2
    while(p * p<= n) : 
        if (n % p == 0) : 
            while (n % p == 0) : 
                n = n // p 
            result = result * (1.0 - (1.0 / (float) (p))) 
        p = p + 1
    if (n > 1) : 
        result = result * (1.0 - (1.0 / (float)(n))) 
   
    return (int)(result) 

def is_prime(n):
	if n == 0:
		return False
	if n == 1:
		return True
	for i in range(2, int(n ** (1 / 2)) + 1):
		if not n % i:
			return False
 
	return True

def next_prime(n, primes):
	while primes[n] != True:
		n += 1
	return n

 
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array
 
################## un-comment below 2 lines when using factorization #################
spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
    res = []
    for i in range(2, int(x ** 0.5) + 1):
    	while x % i == 0:
    		res.append(i)
    		x //= i
    if x != 1:
   		res.append(x)
    return res
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

def factors(n):
	res = []
	for i in range(1, int(n ** 0.5) + 1):
		if n % i == 0:
			res.append(i)
			res.append(n // i)
	return list(set(res))
 
## taking integer array input
def int_array():
    return list(map(int, input().strip().split()));
 
def float_array():
    return list(map(float, input().strip().split()));
 
## taking string array input
def str_array():
    return input().strip().split();

def binary_search(low, high, w, h, n):
	while low < high:
		mid = low + (high - low) // 2
		# print(low, mid, high)
		if check(mid, w, h, n):
			low = mid + 1
		else:
			high = mid
	return low

## for checking any conditions
def check(moves, n):
	val = (moves + 1) // 2
	rem = moves - val
	sol = (val + 1) * (rem + 1)
	return sol < n
	

## for sorting according to second position
def sortSecond(val):
	return val[1]
	

 
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
small = "abcdefghijklmnopqrstuvwxyz"
large = small.upper()

################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
 
from itertools import permutations
import math
import bisect as bis
import random
import sys
import collections as collect
import functools as fnt
from decimal import Decimal
# from sys import stdout
# import numpy as np
"""
_______________
rough work here
_______________

T rooms
every room has a door
each door can be described by
the number n of seals on it
ai is integer energy
n is even
"""

def solve():
    a, b = read()
    if b >= 2 * a:
        print("NO")
    else:
        print("YES")



# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
 
BUFSIZE = 8192
 
 
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()
 
 
if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
 
input = lambda: sys.stdin.readline().rstrip("\r\n")
 
# endregion
 
 
if __name__ == "__main__":
    #read()
    # sys.stdin = open("input.txt", "r")
    # sys.stdout = open("output.txt", "w")
    t = scanner()
    for i in range(t):
    	solve()
    #dmain()
 
# Comment Read()
	# fin_time = datetime.now()
# 	print("Execution time (for loop): ", (fin_time-init_time))
