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>
  #pragma GCC optimize("O3")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
  using namespace std;
  int a, b, c, d, n, m, k, kolt;
  int mas[200002];
  int main() {
    scanf("%d", &kolt);
    for (int _n((kolt)), hod(1); hod <= _n; hod++) {
      scanf("%d", &n);
      c = 0;
      b = 0;
      for (int _n((n)-1), i(0); i <= _n; i++) {
        scanf("%d", &mas[i]);
        c ^= mas[i];
        b += mas[i];
      }
      if (c || b == n) {
        printf("NO\n");
        continue;
      }
      vector<int> res;
      bool bad = 0;
      for (int _n((n)-1), i(0); i <= _n; i++) {
        if (!mas[i]) continue;
        if (i && mas[i - 1] == 1) continue;
        int a = i;
        while (a < n && mas[a] == 1) ++a;
        while ((a - i) % 2 == 1) {
          b = a;
          while (mas[b] == 0) ++b;
          for (int j = b - 2; j >= a; j -= 2) {
            res.push_back(j);
            mas[j] = mas[j + 1] = mas[j + 2] = 1;
            b -= 2;
          }
          if (b == a + 1) {
            res.push_back(a - 1);
            mas[a - 1] = mas[a] = mas[a + 1] = 0;
            --a;
            break;
          } else {
            while (a < n && mas[a] == 1) ++a;
          }
        }
        if (a <= i) continue;
        if (i - 1 >= 0 && mas[i - 1] == 0) {
          for (int j = i; j < a; j += 2) {
            res.push_back(j - 1);
            mas[j - 1] = mas[j] = mas[j + 1] = 0;
          }
        } else if (a < n) {
          for (int j = a - 2; j >= i; j -= 2) {
            res.push_back(j);
            mas[j] = mas[j + 1] = mas[j + 2] = 0;
          }
        } else {
          bad = 1;
          break;
        }
      }
      if (bad) {
        printf("NO\n");
        continue;
      }
      printf("YES\n");
      printf("%d\n", (int)((res).size()));
      if (res.empty()) continue;
      for (int _n(((int)((res).size())) - 1), i(0); i <= _n; i++) {
        if (i) printf(" ");
        printf("%d", res[i] + 1);
      }
      printf("\n");
    }
  }
  