Problem: Phoenix has decided to become a scientist! He is currently investigating the growth of bacteria.

Initially, on day 1, there is one bacterium with mass 1.

Every day, some number of bacteria will split (possibly zero or all). When a bacterium of mass m splits, it becomes two bacteria of mass m/2 each. For example, a bacterium of mass 3 can split into two bacteria of mass 1.5.

Also, every night, the mass of every bacteria will increase by one.

Phoenix is wondering if it is possible for the total mass of all the bacteria to be exactly n. If it is possible, he is interested in the way to obtain that mass using the minimum possible number of nights. Help him become the best scientist!

Input

The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first line of each test case contains an integer n (2 ≤ n ≤ 10^9) — the sum of bacteria masses that Phoenix is interested in. 

Output

For each test case, if there is no way for the bacteria to exactly achieve total mass n, print -1. Otherwise, print two lines.

The first line should contain an integer d — the minimum number of nights needed.

The next line should contain d integers, with the i-th integer representing the number of bacteria that should split on the i-th day.

If there are multiple solutions, print any.

Example

Input


3
9
11
2


Output


3
1 0 2 
3
1 1 2
1
0 

Note

In the first test case, the following process results in bacteria with total mass 9: 

  * Day 1: The bacterium with mass 1 splits. There are now two bacteria with mass 0.5 each. 
  * Night 1: All bacteria's mass increases by one. There are now two bacteria with mass 1.5. 
  * Day 2: None split. 
  * Night 2: There are now two bacteria with mass 2.5. 
  * Day 3: Both bacteria split. There are now four bacteria with mass 1.25. 
  * Night 3: There are now four bacteria with mass 2.25. 

The total mass is 2.25+2.25+2.25+2.25=9. It can be proved that 3 is the minimum number of nights needed. There are also other ways to obtain total mass 9 in 3 nights.

 

In the second test case, the following process results in bacteria with total mass 11: 

  * Day 1: The bacterium with mass 1 splits. There are now two bacteria with mass 0.5. 
  * Night 1: There are now two bacteria with mass 1.5. 
  * Day 2: One bacterium splits. There are now three bacteria with masses 0.75, 0.75, and 1.5. 
  * Night 2: There are now three bacteria with masses 1.75, 1.75, and 2.5. 
  * Day 3: The bacteria with mass 1.75 and the bacteria with mass 2.5 split. There are now five bacteria with masses 0.875, 0.875, 1.25, 1.25, and 1.75. 
  * Night 3: There are now five bacteria with masses 1.875, 1.875, 2.25, 2.25, and 2.75. 

The total mass is 1.875+1.875+2.25+2.25+2.75=11. It can be proved that 3 is the minimum number of nights needed. There are also other ways to obtain total mass 11 in 3 nights.

 

In the third test case, the bacterium does not split on day 1, and then grows to mass 2 during night 1.

Solution (in PYTHON3):

Solution:

from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
#from operator import add,mul,sub,xor,truediv,floordiv
from functools import *

from collections import deque
#------------------------------------------------------------------------
import os
import sys

from io import BytesIO, IOBase
# region fastio
 
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")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
 

#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------


from types import GeneratorType
 
 
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
 
    return wrappedfunc


farr=[1]
ifa=[]
 
def fact(x,mod=0):
    if mod:
        while x>=len(farr):
            farr.append(farr[-1]*len(farr)%mod)
    else:
        while x>=len(farr):
            farr.append(farr[-1]*len(farr))
    return farr[x]
 
def ifact(x,mod):
    global ifa
    fact(x,mod)
    ifa.append(pow(farr[-1],mod-2,mod))
    for i in range(x,0,-1):
        ifa.append(ifa[-1]*i%mod)
    ifa.reverse()
 
def per(i,j,mod=0):
    if i<j: return 0
    if not mod:
        return fact(i)//fact(i-j)
    return farr[i]*ifa[i-j]%mod
    
def com(i,j,mod=0):
    if i<j: return 0
    if not mod:        
        return per(i,j)//fact(j)
    return per(i,j,mod)*ifa[j]%mod
 
def catalan(n):
    return com(2*n,n)//(n+1)
 
def isprime(n):
    for i in range(2,int(n**0.5)+1):
        if n%i==0:
            return False
    return True
 
def lowbit(n):
    return n&-n
 
def inverse(a,m):
    a%=m
    if a<=1: return a
    return ((1-inverse(m,a)*m)//a)%m
 
class BIT:
    def __init__(self,arr):
        self.arr=arr
        self.n=len(arr)-1
        
    def update(self,x,v):
        while x<=self.n:
            self.arr[x]+=v
            x+=x&-x
 
    def query(self,x):
        ans=0
        while x:
            ans+=self.arr[x]
            x&=x-1
        return ans
'''
class SMT:
    def __init__(self,arr):
        self.n=len(arr)-1
        self.arr=[0]*(self.n<<2)
        self.lazy=[0]*(self.n<<2)
        def Build(l,r,rt):
            if l==r:
                self.arr[rt]=arr[l]
                return
            m=(l+r)>>1
            Build(l,m,rt<<1)
            Build(m+1,r,rt<<1|1)
            self.pushup(rt)
        Build(1,self.n,1)
 
    def pushup(self,rt):
        self.arr[rt]=self.arr[rt<<1]+self.arr[rt<<1|1]
 
    def pushdown(self,rt,ln,rn):#lr,rn表区间数字数
        if self.lazy[rt]:
            self.lazy[rt<<1]+=self.lazy[rt]
            self.lazy[rt<<1|1]+=self.lazy[rt]
            self.arr[rt<<1]+=self.lazy[rt]*ln
            self.arr[rt<<1|1]+=self.lazy[rt]*rn
            self.lazy[rt]=0
            
    def update(self,L,R,c,l=1,r=None,rt=1):#L,R表示操作区间
        if r==None: r=self.n
        if L<=l and r<=R:
            self.arr[rt]+=c*(r-l+1)
            self.lazy[rt]+=c
            return
        m=(l+r)>>1
        self.pushdown(rt,m-l+1,r-m)
        if L<=m: self.update(L,R,c,l,m,rt<<1)
        if R>m: self.update(L,R,c,m+1,r,rt<<1|1)
        self.pushup(rt)
 
    def query(self,L,R,l=1,r=None,rt=1):
        if r==None: r=self.n
        #print(L,R,l,r,rt)
        if L<=l and R>=r:
            return self.arr[rt]
        m=(l+r)>>1
        self.pushdown(rt,m-l+1,r-m)
        ans=0
        if L<=m: ans+=self.query(L,R,l,m,rt<<1)
        if R>m: ans+=self.query(L,R,m+1,r,rt<<1|1)
        return ans
'''       
class DSU:#容量+路径压缩
    def __init__(self,n):
        self.c=[-1]*n
 
    def same(self,x,y):
        return self.find(x)==self.find(y)
 
    def find(self,x):
        if self.c[x]<0:
            return x
        self.c[x]=self.find(self.c[x])
        return self.c[x]
 
    def union(self,u,v):
        u,v=self.find(u),self.find(v)
        if u==v:
            return False
        if self.c[u]>self.c[v]:
            u,v=v,u
        self.c[u]+=self.c[v]
        self.c[v]=u
        return True
 
    def size(self,x): return -self.c[self.find(x)]
    
class UFS:#秩+路径
    def __init__(self,n):
        self.parent=[i for i in range(n)]
        self.ranks=[0]*n
 
    def find(self,x):
        if x!=self.parent[x]:
            self.parent[x]=self.find(self.parent[x])
        return self.parent[x]
 
    def union(self,u,v):
        pu,pv=self.find(u),self.find(v)
        if pu==pv:
            return False
        if self.ranks[pu]>=self.ranks[pv]:
            self.parent[pv]=pu
            if self.ranks[pv]==self.ranks[pu]:
                self.ranks[pu]+=1
        else:
            self.parent[pu]=pv
 
def Prime(n):
    c=0
    prime=[]
    flag=[0]*(n+1)
    
    for i in range(2,n+1):
        if not flag[i]:
            prime.append(i)
            c+=1
        for j in range(c):
            if i*prime[j]>n: break
            flag[i*prime[j]]=prime[j]
            if i%prime[j]==0: break
    return prime
 
def dij(s,graph):
    d={}
    d[s]=0
    heap=[(0,s)]
    seen=set()
    while heap:
        dis,u=heappop(heap)
        if u in seen:
            continue
        seen.add(u)
        for v,w in graph[u]:
            if v not in d or d[v]>d[u]+w:
                d[v]=d[u]+w
                heappush(heap,(d[v],v))
    return d
 
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
def lcm(a,b): return a*b//gcd(a,b) 
 
class DLN:
    def __init__(self,val):
        self.val=val
        self.pre=None
        self.next=None

def nb(i,j):
    for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
        if 0<=ni<n and 0<=nj<m:
            yield ni,nj
            
@bootstrap
def gdfs(r,p):
    if len(g[r])==1 and p!=-1:
        yield None
    for ch in g[r]:
        if ch!=p:
            yield gdfs(ch,r)
    yield None



def topo(n):
    q=deque()
    res=[]
    for i in range(1,n+1):
        if ind[i]==0:
            q.append(i)
            res.append(i)
    while q:
        u=q.popleft()
        for v in g[u]:
            ind[v]-=1
            if ind[v]==0:
                q.append(v)
                res.append(v)
    return res

t=N()
for i in range(t):
    n=N()
    res=[2]
    cur=2
    w=3
    while w<n:
        cur*=2
        res.append(cur)
        w+=cur
    ans=len(res)
    j=len(res)-1
    #print(w,res)
    while w>n:
        if j:
            if w-n<=res[j-1]:
                res[j]-=w-n
                w=n
            else:
                res[j]=res[j-1]
                w-=res[j-1]
                j-=1
        else:
            res[j]-=1
            w-=1
        #print(w,res)
    #print(res)
    pre=1
    for i in range(len(res)):
        pre,res[i]=res[i],res[i]-pre
    print(ans)
    print(*res)
            
            
        
        
            
            
        
 


''' 
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''

''' 
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''