description: You are given a sequence a of length n consisting of 0s and 1s.
  
  You can perform the following operation on this sequence: 
  
    * Pick an index i from 1 to n-2 (inclusive). 
    * Change all of a_{i}, a_{i+1}, a_{i+2} to a_{i} ⊕ a_{i+1} ⊕ a_{i+2} simultaneously, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
  
  Find a sequence of at most n operations that changes all elements of a to 0s or report that it's impossible.
  
  We can prove that if there exists a sequence of operations of any length that changes all elements of a to 0s, then there is also such a sequence of length not greater than n.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4).
  
  The first line of each test case contains a single integer n (3 ≤ n ≤ 2⋅10^5) — the length of a.
  
  The second line of each test case contains n integers a_1, a_2, …, a_n (a_i = 0 or a_i = 1) — elements of a.
  
  It is guaranteed that the sum of n over all test cases does not exceed 2⋅10^5.
  
  Output
  
  For each test case, do the following: 
  
    * if there is no way of making all the elements of a equal to 0 after performing the above operation some number of times, print "NO". 
    * otherwise, in the first line print "YES", in the second line print k (0 ≤ k ≤ n) — the number of operations that you want to perform on a, and in the third line print a sequence b_1, b_2, ..., b_k (1 ≤ b_i ≤ n - 2) — the indices on which the operation should be applied. 
  
  
  
  If there are multiple solutions, you may print any.
  
  Example
  
  Input
  
  
  3
  3
  0 0 0
  5
  1 1 1 1 0
  4
  1 0 0 1
  
  
  Output
  
  
  YES
  0
  YES
  2
  3 1
  NO
  
  Note
  
  In the first example, the sequence contains only 0s so we don't need to change anything.
  
  In the second example, we can transform [1, 1, 1, 1, 0] to [1, 1, 0, 0, 0] and then to [0, 0, 0, 0, 0] by performing the operation on the third element of a and then on the first element of a.
  
  In the third example, no matter whether we first perform the operation on the first or on the second element of a we will get [1, 1, 1, 1], which cannot be transformed to [0, 0, 0, 0].
solution: #include <bits/stdc++.h>
  using namespace std;
  vector<long long> solveoddarray(vector<long long>& a, long long start,
                                  long long end) {
    vector<long long> ans;
    for (long long i = start; i <= end - 2; i += 2) {
      ans.push_back(i + 1);
    }
    for (long long i = end - 4; i >= start; i -= 2) {
      ans.push_back(i + 1);
    }
    return ans;
  }
  int32_t main() {
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      scanf("%lld", &n);
      vector<long long> a(n);
      long long sum = 0;
      for (long long i = 0; i < n; i++) {
        scanf("%lld", &a[i]);
        sum += a[i];
      }
      bool possible = true;
      vector<long long> ans;
      if (sum % 2 == 0) {
        if (n % 2 == 1) {
          ans = solveoddarray(a, 0, n - 1);
        } else {
          if (a[0] == 0) {
            ans = solveoddarray(a, 1, n - 1);
          } else {
            long long s = 0;
            possible = false;
            for (long long i = 0; i < n; i++) {
              s += a[i];
              vector<long long> t1;
              if (s % 2 == 0 && i % 2 == 0) {
                ans = solveoddarray(a, 0, i);
                t1 = solveoddarray(a, i + 1, n - 1);
                for (long long j = 0; j < (long long)t1.size(); j++) {
                  ans.push_back(t1[j]);
                }
                possible = true;
                break;
              }
            }
          }
        }
      } else {
        possible = false;
      }
      if (!possible) {
        printf("NO\n");
      } else {
        printf("YES\n");
        printf("%lld\n", (long long)ans.size());
        if ((long long)ans.size() > 0) {
          for (long long i = 0; i < (long long)ans.size(); i++) {
            printf("%lld ", ans[i]);
          }
          printf("\n");
        }
      }
    }
  }
  