description: 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.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int oo = 1000000099;
  const int mod = 1000000007;
  const int maxn = 1000011;
  int ntest;
  int n, a[maxn];
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    if (fopen(""
              ".inp",
              "r")) {
      freopen(
          ""
          ".inp",
          "r", stdin);
      freopen(
          ""
          ".out",
          "w", stdout);
    }
    cin >> ntest;
    while (ntest--) {
      int n;
      cin >> n;
      for (int i = (1); i <= (n); ++i) cin >> a[i];
      if (n % 2 == 0)
        cout << "YES\n";
      else {
        int cur = 0;
        for (int i = (1); i <= (n - 1); ++i) {
          if (a[i] >= a[i + 1]) cur = 1;
        }
        if (cur)
          cout << "YES\n";
        else
          cout << "NO\n";
      }
    }
    return 0;
  }
  