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;
  template <typename... Args>
  void trace(Args... args) {}
  long long t = 1, n, ans, a[1001000];
  int main() {
    scanf("%lld", &t);
  st:
    ans = 0;
    scanf("%lld", &n);
    long long xo = 0, f[] = {0, 0};
    vector<int> ones;
    for (int i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
      f[a[i]]++;
      xo ^= a[i];
      if (a[i]) ones.push_back(i);
    }
    if (xo == 1 || f[0] == 0)
      puts("NO");
    else {
      vector<int> op;
      int l = 0, r = 0;
      while (1) {
        while (r < n) {
          if (a[l] == 0) {
            l++;
            r = l;
          } else if (a[r] == 1) {
            r++;
          } else
            break;
        }
        if (l == n)
          break;
        else if (l == 0 && r == n)
          break;
        else if ((r - l) % 2 == 0) {
          if (l > 0) {
            while (l < r) {
              op.push_back(l);
              assert(a[l - 1] == 0 && a[l] == 1 && a[l + 1] == 1);
              a[l - 1] = a[l] = a[l + 1] = 0;
              l += 2;
            }
          } else {
            for (int j = r; j > l; j -= 2) {
              op.push_back(j - 1);
              assert(a[j - 2] == 1 && a[j - 1] == 1 && a[j] == 0);
              a[j - 2] = a[j - 1] = a[j] = 0;
            }
          }
        } else {
          if (r + 1 >= n) break;
          long long tmp = a[r - 1] ^ a[r] ^ a[r + 1];
          a[r - 1] = a[r] = a[r + 1] = tmp;
          op.push_back(r);
          if (tmp == 0) r--;
        }
      }
      int can = 1;
      for (int i = 0; i < n; i++) can &= (a[i] == 0);
      if (!can)
        puts("NO");
      else {
        printf("YES\n%d\n", (int)op.size());
        for (auto x : op) printf("%d ", x);
        puts("");
      }
    }
    if (--t) goto st;
    return 0;
  }
  