YouKn0wWho has an integer sequence a_1, a_2, … a_n. Now he will split the sequence a into one or more consecutive subarrays so that each element of a belongs to exactly one subarray. Let k be the number of resulting subarrays, and h_1, h_2, …, h_k be the lengths of the longest increasing subsequences of corresponding subarrays.

For example, if we split [2, 5, 3, 1, 4, 3, 2, 2, 5, 1] into [2, 5, 3, 1, 4], [3, 2, 2, 5], [1], then h = [3, 2, 1].

YouKn0wWho wonders if it is possible to split the sequence a in such a way that the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of h_1, h_2, …, h_k is equal to 0. You have to tell whether it is possible.

The longest increasing subsequence (LIS) of a sequence b_1, b_2, …, b_m is the longest sequence of valid indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and b_{i_1} < b_{i_2} < … < b_{i_k}. For example, the LIS of [2, 5, 3, 3, 5] is [2, 3, 5], which has length 3.

An array c is a subarray of an array b if c can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.

Input

The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5).

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).

It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.

Output

For each test case, print "YES" (without quotes) if it is possible to split into subarrays in the desired way, print "NO" (without quotes) otherwise. You can print each letter in any register (upper or lower).

Example

Input


4
7
1 3 4 2 2 1 5
3
1 3 4
5
1 3 2 4 2
4
4 3 2 1


Output


YES
NO
YES
YES

Note

In the first test case, YouKn0wWho can split the sequence in the following way: [1, 3, 4], [2, 2], [1, 5]. This way, the LIS lengths are h = [3, 1, 2], and the bitwise XOR of the LIS lengths is 3 ⊕ 1 ⊕ 2 = 0.

In the second test case, it can be shown that it is impossible to split the sequence into subarrays that will satisfy the condition.
import os
import sys
from io import BytesIO, IOBase
 
_str = str
str = lambda x=b"": x if type(x) is bytes else _str(x).encode()
 
BUFSIZE = 8192
 
 
def sigma1(a):
    return sum(a)
def sigma2(a):
    return sum([i*i for i in a])
def sigma3(a):
    return sum([i*i*i for i in a])
def prodofpairs(a):
    return (sigma1(a)**2-sigma2(a))//2
def prodoftriplets(a):
    return (sigma1(a)**3-3*sigma1(a)*sigma2(a)+2*sigma3(a))//6
from collections import Counter,defaultdict
from math import gcd
from collections import deque
def bfs(graph, start=0):
    used = [False] * len(graph)
    used[start] = True
    q = [start]
    for v in q:
        for w in graph[v]:
            if not used[w]:
                used[w] = True
                q.append(w)
 
 
def layers(graph, start=0):
    used = [False] * len(graph)
    used[start] = True
    q, ret = [start], []
    level=[0]*len(graph)
    while q:
        nq = []
        ret.append(q)
        for v in q:
            for w in graph[v]:
                if not used[w]:
                    used[w] = True
                    nq.append(w)
                    level[w]=1+level[v]
        q = nq
    return level
 
import math
 
# function to find GCD of sub sequence
# of size k with max GCD in the array
def findMaxGCD(arr, n, k):
    high = max(arr)
    divisors = [0] * (high + 1)
    for i in range(n) :
        for j in range(1, int(math.sqrt(arr[i])) + 1):
            if (arr[i] % j == 0) :
                divisors[j] += 1
                if (j != arr[i] // j):
                    divisors[arr[i] // j] += 1
    for i in range(high, 0, -1):
        if (divisors[i] >= k):
            return i
def xdist(a,b):
    if(b[0]>=a[0]>=b[2] or b[0]<=a[0]<=b[2] ):
        return 0
    return min(abs(a[0]-b[0]),abs(a[0]-b[2]))
def ydist(a,b):
    if(b[1]>=a[1]>=b[3] or b[1]<=a[1]<=b[3] ):
        return 0
    return min(abs(a[1]-b[1]),abs(a[1]-b[3]))
def countdistx(i,j,k,obj):
    m=0
    for i1 in range(k):
        m+=xdist([i,j],obj[i1])
    return m
def countdisty(i,j,k,obj):
    m=0
    for i1 in range(k):
        m+=ydist([i,j],obj[i1])
    return m
from bisect import bisect_left
def dp(i,n,cost,a):
    if(cost==0):
        return 0
    if(cost<0):
        return float("Inf")
    l=[float("Inf")]
    for j in range(i,n+1):
        k=bisect_left(a,a[j]+cost,j)
        if(k<=n and a[k]==a[j]+cost):
            l.append(k-j)
    return min(l)
def isp(x):
    for i in range(2,int(math.sqrt(x))+1):
        if(x%i==0):
            return False
    return True
from bisect import bisect_right
from sys import stdout
from math import ceil,log2,floor
def main():
    for _ in range(int(input())):
        n=int(input())
        a=list(map(int,input().split()))
        if(n%2==0):
           print("YES")
        else:
            for i in range(n-1):
                if(a[i]>=a[i+1]):
                    print("YES")
                    break
            else:
                print("NO")
    



 
    
 
 
 
            
 
 
 
    
        
        
            
 
 
 
 
        
 
 
 
 
 
        
 
 
        
 
main()
 
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")
 
 
 