Problem: Recently, Olya received a magical square with the size of 2^n× 2^n.

It seems to her sister that one square is boring. Therefore, she asked Olya to perform exactly k splitting operations.

A Splitting operation is an operation during which Olya takes a square with side a and cuts it into 4 equal squares with side a/2. If the side of the square is equal to 1, then it is impossible to apply a splitting operation to it (see examples for better understanding).

Olya is happy to fulfill her sister's request, but she also wants the condition of Olya's happiness to be satisfied after all operations.

The condition of Olya's happiness will be satisfied if the following statement is fulfilled:

Let the length of the side of the lower left square be equal to a, then the length of the side of the right upper square should also be equal to a. There should also be a path between them that consists only of squares with the side of length a. All consecutive squares on a path should have a common side.

Obviously, as long as we have one square, these conditions are met. So Olya is ready to fulfill her sister's request only under the condition that she is satisfied too. Tell her: is it possible to perform exactly k splitting operations in a certain order so that the condition of Olya's happiness is satisfied? If it is possible, tell also the size of the side of squares of which the path from the lower left square to the upper right one will consist.

Input

The first line contains one integer t (1 ≤ t ≤ 10^3) — the number of tests.

Each of the following t lines contains two integers n_i and k_i (1 ≤ n_i ≤ 10^9, 1 ≤ k_i ≤ 10^{18}) — the description of the i-th test, which means that initially Olya's square has size of 2^{n_i}× 2^{n_i} and Olya's sister asks her to do exactly k_i splitting operations.

Output

Print t lines, where in the i-th line you should output "YES" if it is possible to perform k_i splitting operations in the i-th test in such a way that the condition of Olya's happiness is satisfied or print "NO" otherwise. If you printed "YES", then also print the log_2 of the length of the side of the squares through space, along which you can build a path from the lower left square to the upper right one.

You can output each letter in any case (lower or upper).

If there are multiple answers, print any.

Example

Input


3
1 1
2 2
2 12


Output


YES 0
YES 1
NO

Note

In each of the illustrations, the pictures are shown in order in which Olya applied the operations. The recently-created squares are highlighted with red.

In the first test, Olya can apply splitting operations in the following order:

<image> Olya applies one operation on the only existing square.

The condition of Olya's happiness will be met, since there is a path of squares of the same size from the lower left square to the upper right one:

<image>

The length of the sides of the squares on the path is 1. log_2(1) = 0.

In the second test, Olya can apply splitting operations in the following order:

<image> Olya applies the first operation on the only existing square. She applies the second one on the right bottom square.

The condition of Olya's happiness will be met, since there is a path of squares of the same size from the lower left square to the upper right one:

<image>

The length of the sides of the squares on the path is 2. log_2(2) = 1.

In the third test, it takes 5 operations for Olya to make the square look like this:

<image>

Since it requires her to perform 7 splitting operations, and it is impossible to perform them on squares with side equal to 1, then Olya cannot do anything more and the answer is "NO".

Solution (in PYTHON3):

Solution:
''' CODED WITH LOVE BY SATYAM KUMAR '''


from sys import stdin, stdout
import cProfile, math
from collections import Counter
from bisect import bisect_left,bisect,bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
import sys, threading
sys.setrecursionlimit(10**6) # max depth of recursion
threading.stack_size(2**27)  # new thread will get stack of such size

printHeap = str()
memory_constrained = False
P = 10**9+7
import sys
sys.setrecursionlimit(10000000)

class Operation:
    def __init__(self, name, function, function_on_equal, neutral_value=0):
        self.name = name
        self.f = function
        self.f_on_equal = function_on_equal
def add_multiple(x, count):
    return x * count
def min_multiple(x, count):
    return x
def max_multiple(x, count):
    return x
sum_operation = Operation("sum", sum, add_multiple, 0)
min_operation = Operation("min", min, min_multiple, 1e9)
max_operation = Operation("max", max, max_multiple, -1e9)
class SegmentTree:
    def __init__(self,
                 array,
                 operations=[sum_operation, min_operation, max_operation]):
        self.array = array
        if type(operations) != list:
            raise TypeError("operations must be a list")
        self.operations = {}
        for op in operations:
            self.operations[op.name] = op
        self.root = SegmentTreeNode(0, len(array) - 1, self)
    def query(self, start, end, operation_name):
        if self.operations.get(operation_name) == None:
            raise Exception("This operation is not available")
        return self.root._query(start, end, self.operations[operation_name])
    def summary(self):
        return self.root.values
    def update(self, position, value):
        self.root._update(position, value)
    def update_range(self, start, end, value):
        self.root._update_range(start, end, value)
    def __repr__(self):
        return self.root.__repr__()
class SegmentTreeNode:
    def __init__(self, start, end, segment_tree):
        self.range = (start, end)
        self.parent_tree = segment_tree
        self.range_value = None
        self.values = {}
        self.left = None
        self.right = None
        if start == end:
            self._sync()
            return
        self.left = SegmentTreeNode(start, start + (end - start) // 2,
                                    segment_tree)
        self.right = SegmentTreeNode(start + (end - start) // 2 + 1, end,
                                     segment_tree)
        self._sync()
    def _query(self, start, end, operation):
        if end < self.range[0] or start > self.range[1]:
            return None
        if start <= self.range[0] and self.range[1] <= end:
            return self.values[operation.name]
        self._push()
        left_res = self.left._query(start, end,
                                    operation) if self.left else None
        right_res = self.right._query(start, end,
                                      operation) if self.right else None
        if left_res is None:
            return right_res
        if right_res is None:
            return left_res
        return operation.f([left_res, right_res])
    def _update(self, position, value):
        if position < self.range[0] or position > self.range[1]:
            return
        if position == self.range[0] and self.range[1] == position:
            self.parent_tree.array[position] = value
            self._sync()
            return
        self._push()
        self.left._update(position, value)
        self.right._update(position, value)
        self._sync()
    def _update_range(self, start, end, value):
        if end < self.range[0] or start > self.range[1]:
            return
        if start <= self.range[0] and self.range[1] <= end:
            self.range_value = value
            self._sync()
            return
        self._push()
        self.left._update_range(start, end, value)
        self.right._update_range(start, end, value)
        self._sync()
    def _sync(self):
        if self.range[0] == self.range[1]:
            for op in self.parent_tree.operations.values():
                current_value = self.parent_tree.array[self.range[0]]
                if self.range_value is not None:
                    current_value = self.range_value
                self.values[op.name] = op.f([current_value])
        else:
            for op in self.parent_tree.operations.values():
                result = op.f(
                    [self.left.values[op.name], self.right.values[op.name]])
                if self.range_value is not None:
                    bound_length = self.range[1] - self.range[0] + 1
                    result = op.f_on_equal(self.range_value, bound_length)
                self.values[op.name] = result
    def _push(self):
        if self.range_value is None:
            return
        if self.left:
            self.left.range_value = self.range_value
            self.right.range_value = self.range_value
            self.left._sync()
            self.right._sync()
            self.range_value = None
    def __repr__(self):
        ans = "({}, {}): {}\n".format(self.range[0], self.range[1],
                                      self.values)
        if self.left:
            ans += self.left.__repr__()
        if self.right:
            ans += self.right.__repr__()
        return ans

def display(string_to_print):
    stdout.write(str(string_to_print) + "\n")

def primeFactors(n): #n**0.5 complex 
    factors = dict()
    for i in range(2,math.ceil(math.sqrt(n))+1):  
        while n % i== 0: 
            if i in factors:
                factors[i]+=1
            else: factors[i]=1
            n = n // i 
    if n>2:
        factors[n]=1
    return (factors)
    
def isprime(n):
    """Returns True if n is prime."""
    if n < 4:
        return True
    if n % 2 == 0:
        return False
    if n % 3 == 0:
        return False
    i = 5
    w = 2
    while i * i <= n:
        if n % i == 0:
            return False
        i += w
        w = 6 - w
    return True

def test_print(*args):
    if testingMode:
        print(args)

def display_list(list1, sep=" "):
    stdout.write(sep.join(map(str, list1)) + "\n")

def get_int():
    return int(stdin.readline().strip())

def get_tuple():
    return map(int, stdin.readline().split())

def get_list():
    return list(map(int, stdin.readline().split()))

memory = dict()
def clear_cache():
    global memory
    memory = dict()
def cached_fn(fn, *args):
    global memory
    if args in memory:
        return memory[args]
    else:
        result = fn(*args)
        memory[args] = result
        return result


# -------------------------------------------------------------- MAIN PROGRAM
TestCases = True
testingMode = False
optimiseForReccursion = False #Can not be used clubbed with TestCases
            

def main():
    n,k= get_tuple()
    x = 0
    i = 0
    xs=[]
    if (k==0):
        print("YES "+str(n))
        return
    if (n==1):
        print("YES 0") if k==1 else print("NO")
        return
    while x<=k and i<n:
        x+=4**i
        xs.append(x)
        i+=1
   # print(xs)
    if x<k:
        print("NO")
        return
    elif k==3:
        print("YES "+str((n-1))) if n>2 else print("NO")
    else:
        for i,ele in enumerate(xs):
            if i==len(xs)-1:
                ind = len(xs)-1
              #  print(ind,"At end")
            else:
                if ele<=k and xs[i+1]>k:
                    ind = i+1
                   # print(ind,"inBW")
                    break
    
        #print(xs[ind-1]+(2*(2**(ind)-2)+3))
        if (xs[ind-1]+(2*(2**(ind)-2)+3)<=k) and n>1:
            ind+=1
        #print(ind,n)
        print("YES "+str((n-ind)))



    
        


        
            


# --------------------------------------------------------------------- END


if TestCases: 
    for _ in range(get_int()): 
        cProfile.run('main()') if testingMode else main() 
else: (cProfile.run('main()') if testingMode else main()) if not optimiseForReccursion else threading.Thread(target=main).start()