human
Petya has an array of integers a_1, a_2, …, a_n. He only likes sorted arrays. Unfortunately, the given array could be arbitrary, so Petya wants to sort it.

Petya likes to challenge himself, so he wants to sort array using only 3-cycles. More formally, in one operation he can pick 3 pairwise distinct indices i, j, and k (1 ≤ i, j, k ≤ n) and apply i → j → k → i cycle to the array a. It simultaneously places a_i on position j, a_j on position k, and a_k on position i, without changing any other element.

For example, if a is [10, 50, 20, 30, 40, 60] and he chooses i = 2, j = 1, k = 5, then the array becomes [\underline{50}, \underline{40}, 20, 30, \underline{10}, 60].

Petya can apply arbitrary number of 3-cycles (possibly, zero). You are to determine if Petya can sort his array a, i. e. make it non-decreasing.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5 ⋅ 10^5). Description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the length of the array a.

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

It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5.

Output

For each test case, print "YES" (without quotes) if Petya can sort the array a using 3-cycles, and "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower).

Example

Input

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

Output

YES
YES
NO
YES
NO
YES
YES

Note

In the 6-th test case Petya can use the 3-cycle 1 → 3 → 2 → 1 to sort the array.

In the 7-th test case Petya can apply 1 → 3 → 2 → 1 and make a = [1, 4, 2, 3]. Then he can apply 2 → 4 → 3 → 2 and finally sort the array.
Solve the task in PYTHON3.

=====
Give me a Python program to determine if Petya can sort his array a, i?
assistant
#Here is Python program to determine if Petya can sort his array a, i.

# cook your dish here
import sys
import bisect
from bisect import bisect_left as lb
from bisect import bisect_right as rb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
sflo=lambda:list(map(float,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
sys.setrecursionlimit(3*10**5)
mod=10**9+7
mod1=998244353
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs2(root,par):
    if par!=-1:
        dp[root]=dp[par]+1
    for i in range(1,20):
        if kthpar[root][i-1]!=-1:
            kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
    for child in gp[root]:
        if child==par:continue
        kthpar[child][0]=root
        dfs(child,root)
        
ans=0
a=[]
n,k=0,0
b=[]
vis=[]
tot=0
time=[]
time1=[]
adj=[]
mx=-1
eps=0.0000001
x=0
pref=[]
lvl=[]
gp=[]
ans=[]
f=0
w=[]
#Fenwick Tree point update range sum
def update_it(bit,i,val):
    n=len(bit)
    while(i<n):
        bit[i]+=val
        i+=i&(-i)
def get_ans(bit,i):
    n=len(bit)
    tot=0
    while(i>0):
        tot+=bit[i]
        i-=i&(-i)
    return tot
class Disjoint_SU:
    def __init__(self,n):
        self.parent=[i for i in range(n)]
        self.size=[1 for i in range(n)]
    #find
    def lake_de(self,a):
        if a==self.parent[a]:
            return a
        self.parent[a]=self.lake_de(self.parent[a])
        return self.parent[a]
    #union
    #by size
    def jod(self,a,b):
        x=self.lake_de(a)
        y=self.lake_de(b)
        #for cycle in graph
        if x==y:
            return True
        if self.size[x]>self.size[y]:
            self.size[x]+=self.size[y]
            self.parent[y]=x
        else:
            self.size[y]+=self.size[x]
            self.parent[x]=y
        return False

def dfs(i,prod,rem,n):
    if rem==0:
       ...