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;
  const int maxn = 200005;
  int T, n, m;
  int a[maxn], ans[maxn];
  bool solve(int p, bool b) {
    if (p > n) return true;
    if (!a[p]) return solve(p + 1, true);
    int q = p + 1;
    while (q <= n && !a[q]) q++;
    if (q > n) return false;
    int t = p;
    while (t + 2 < q) ans[++m] = t, t += 2;
    if (t + 1 == q) {
      if (b) {
        for (int i = p; i <= t; i += 2) ans[++m] = i - 1;
        if (!solve(q + 1, true)) return false;
      } else {
        if (q == n || !solve(q + 1, false)) return false;
        for (int i = t; i >= p; i -= 2) ans[++m] = i;
      }
    } else {
      for (int i = t; i >= p; i -= 2) ans[++m] = i;
      if (!solve(q + 1, true)) return false;
    }
    return true;
  }
  int main() {
    scanf("%d", &T);
    while (T--) {
      scanf("%d", &n);
      for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
      m = 0;
      bool res = solve(1, false);
      if (!res)
        puts("NO");
      else {
        puts("YES");
        printf("%d\n", m);
        for (int i = 1; i <= m; i++) printf("%d%c", ans[i], i == m ? '\n' : ' ');
      }
    }
    return 0;
  }
  