Problem: This is an easy version of the problem. In this version, all numbers in the given array are distinct and the constraints on n are less than in the hard version of the problem.

You are given an array a of n integers (there are no equals elements in the array). You can perform the following operations on array elements:

  1. choose any index i (1 ≤ i ≤ n) and move the element a[i] to the begin of the array; 
  2. choose any index i (1 ≤ i ≤ n) and move the element a[i] to the end of the array. 



For example, if n = 5, a = [4, 7, 2, 3, 9], then the following sequence of operations can be performed: 

  * after performing the operation of the first type to the second element, the array a will become [7, 4, 2, 3, 9]; 
  * after performing the operation of the second type to the second element, the array a will become [7, 2, 3, 9, 4]. 



You can perform operations of any type any number of times in any order.

Find the minimum total number of operations of the first and second type that will make the a array sorted in non-decreasing order. In other words, what is the minimum number of operations that must be performed so the array satisfies the inequalities a[1] ≤ a[2] ≤ … ≤ a[n].

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.

Each test case starts with a line containing an integer n (1 ≤ n ≤ 3000) — length of the array a.

Then follow n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9) — an array that needs to be sorted by the given operations. All numbers in the given array are distinct.

The sum of n for all test cases in one test does not exceed 3000.

Output

For each test case output one integer — the minimum total number of operations of the first and second type, which will make the array sorted in non-decreasing order.

Example

Input


4
5
4 7 2 3 9
5
3 5 8 1 7
5
1 4 5 7 12
4
0 2 1 3


Output


2
2
0
2

Note

In the first test case, you first need to move 3, and then 2 to the beginning of the array. Therefore, the desired sequence of operations: [4, 7, 2, 3, 9] → [3, 4, 7, 2, 9] → [2, 3, 4, 7, 9].

In the second test case, you need to move the 1 to the beginning of the array, and the 8 — to the end. Therefore, the desired sequence of operations: [3, 5, 8, 1, 7] → [1, 3, 5, 8, 7] → [1, 3, 5, 7, 8].

In the third test case, the array is already sorted.

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 *

#------------------------------------------------------------------------
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

mod=10**9+7
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 floorsum(a,b,c,n):#sum((a*i+b)//c for i in range(n+1))
    if a==0:return b//c*(n+1)
    if a>=c or b>=c: return floorsum(a%c,b%c,c,n)+b//c*(n+1)+a//c*n*(n+1)//2
    m=(a*n+b)//c
    return n*m-floorsum(c,c-b-1,a,m-1)
 
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 flag,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)
def lis(nums):
    res=[]
    for k in nums:
        i=bisect.bisect_left(res,k)
        if i==len(res):
            res.append(k)
        else:
            res[i]=k
    return len(res)
 
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

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

@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

@bootstrap
def dfs(r):
    vis[r]=1
    ans[r-1]=val[r]
    for ch in g[r]:
        if d[ch]>d[r]:
            yield dfs(ch)
            ans[r-1]=min(ans[r-1],ans[ch-1])
    yield None

def f(x,g):
    ans=0
    for ch in res:
        ans+=c[ch]//x
    return ans>=g

t=N()
for i in range(t):
    n=N()
    a=RLL()
    d={x:i for i,x in enumerate(a)}
    res=sorted(a)
    b=[0]*(n-1)
    for i in range(n-1):
        if d[res[i]]<d[res[i+1]]:
            b[i]=1
    g=GP(b)
    ans=0
    for ch,l in g:
        if ch==1:
            ans=max(ans,l)
    ans=n-ans-1
    print(ans)
    
            
        
        
 


''' 
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()
'''