human
PYTHON3 solution for "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.":


 Write 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.
 class BIT:

    def __init__(self, L):
        self.N = len(L)
        self.bit = [0]*self.N
        for i,l in enumerate(L):
            self.add(i,l)
        self.N0 = 1
        while self.N0*2 <= self.N:
            self.N0 *= 2

    def add(self, a, w):
        x = a + 1
        for i in range(1000):
            self.bit[x-1] += w
            x += x & -x
            if x > self.N:
                break

    def sum(self, a):
        x = a+1
        ret = 0
        for i in range(1000):
            ret += self.bit[x-1]
            x -= x & -x
            if x <= 0:
                break
        return ret

    #you can use this function when the BIT has only non-negative values.
    def lower_bound(self, w):
        if w<=0:
            return 0
        x = 0
        k = self.N0
        while k>0:
            if x+k<=self.N:
                if self.bit[x+k-1]<w:
                    w-=self.bit[x+k-1]
                    x+=k
            k//=2
        return x+1

def solve():
    N = int(input())
    A = list(map(int,input().split()))
    if len(set(A))<len(A):
        print("YES")
        return
    bit = BIT([0]*N)
    A = [(a,i) for i,a in enumerate(A)]
    A.sort(key=lambda x:x[0], reverse=True)
    tento = 0
    buf = []
    prev = A[0][0]
    for a,i in A:
        if prev != a:
            for b in buf:
                bit.add(b, 1)
            buf = []
            prev = a
        tento += bit.sum(i)
        buf.append(i)
    if tento%2==0:
        print("YES")
    else:
        print("NO")

T = int(input())

for i in range(T):
    solve()