XOR Specia-LIS-t

YouKn0wWho has an integer sequence a1,a2,…ana1,a2,…an. Now he will split the sequence aa into one or more consecutive subarrays so that each element of aa belongs to exactly one subarray. Let kk be the number of resulting subarrays, and h1,h2,…,hkh1,h2,…,hk 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][2,5,3,1,4,3,2,2,5,1] into [2,5,3,1,4][2,5,3,1,4], [3,2,2,5][3,2,2,5], [1][1], then h=[3,2,1]h=[3,2,1].

YouKn0wWho wonders if it is possible to split the sequence aa in such a way that the bitwise XOR of h1,h2,…,hkh1,h2,…,hk is equal to 00. You have to tell whether it is possible.

The longest increasing subsequence (LIS) of a sequence b1,b2,…,bmb1,b2,…,bm is the longest sequence of valid indices i1,i2,…,iki1,i2,…,ik such that i1<i2<…<iki1<i2<…<ik and bi1<bi2<…<bikbi1<bi2<…<bik. For example, the LIS of [2,5,3,3,5][2,5,3,3,5] is [2,3,5][2,3,5], which has length 33.

An array cc is a subarray of an array bb if cc can be obtained from bb 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 tt (1≤t≤100001≤t≤10000) — the number of test cases.

The first line of each test case contains a single integer nn (2≤n≤1052≤n≤105).

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109).

It is guaranteed that the sum of nn over all test cases doesn't exceed 3⋅1053⋅105.

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

Copy

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

Copy

YES
NO
YES
YES

Note

In the first test case, YouKn0wWho can split the sequence in the following way: [1,3,4][1,3,4], [2,2][2,2], [1,5][1,5]. This way, the LIS lengths are h=[3,1,2]h=[3,1,2], and the bitwise XOR of the LIS lengths is 3⊕1⊕2=03⊕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.

思路：要想异或为0那就可以令它们相等，这样异或就为0了。如何让它们相等呢？因为此时还需要考虑最大上升子序列，取的时候会非常麻烦，所以我们直接去1个，这样我们就不用考虑大小了。

此时就要分奇偶了，偶直接就能1异或1了，直接为0；开始考虑奇了，奇数不为1时就一定有3个能取（去掉偶数个），上升子序列最小也要是1，所得两个数抵消，三个数分成两个，就一定有一个是1，一个是2个，一的一定是1，那么2的也要是1，为1只有递减，所以就有了答案。

#include <iostream>using namespace std;int main()
{int t;cin>>t;while(t--){int n;cin>>n;int a[n];bool flag=false;if(n%2==0) flag=true;cin>>a[0];for(int i=1;i<n;i++){cin>>a[i];if(a[i]<=a[i-1]){flag=true;}}if(flag) cout<<"YES"<<endl;else cout<<"NO"<<endl;}return 0;
}

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