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 N = 1e6 + 10;
  const long long mod = 1e9 + 7;
  inline long long gcd(long long a, long long b) {
    return !b ? a : gcd(b, a % b);
  }
  inline long long q_pow(long long a, long long x = mod - 2) {
    long long ans = 1, tmp = a;
    while (x) {
      if (x & 1) (ans *= tmp) %= mod;
      (tmp *= tmp) %= mod;
      x >>= 1;
    }
    return ans;
  }
  template <typename T>
  inline void re(T &N) {
    int f = 1;
    char c;
    while ((c = getchar()) < '0' || c > '9')
      if (c == '-') f = -1;
    N = c - '0';
    while ((c = getchar()) >= '0' && c <= '9') N = N * 10 + c - '0';
    N *= f;
  }
  template <class T, class... T_>
  inline void re(T &x, T_ &...y) {
    re(x), re(y...);
  }
  int m, n, t = 1, st, en;
  int a[N], b[N];
  char s[N];
  int main() {
    re(t);
    while (t--) {
      re(n);
      int now = 0;
      for (int i = 1; i <= n; i++) re(a[i]), now ^= a[i];
      if (now) {
        puts("NO");
        continue;
      }
      vector<int> ans;
      int ok = 1;
      for (int i = 1; i <= n && ok; i++)
        if (a[i]) {
          int j = i;
          while (j <= n && a[j]) j++;
          int cnt = j - i;
          while (cnt & 1) {
            while (j <= n && !a[j]) {
              if (j == n) {
                ok = 0;
                break;
              }
              if (!a[j + 1])
                ans.push_back(j - 1), a[j] = a[j + 1] = 1, j += 2;
              else {
                ans.push_back(j - 1), a[j - 1] = a[j] = a[j + 1] = 0;
                j--;
                break;
              }
            }
            if (!ok) break;
            if (a[j]) {
              while (j <= n && a[j]) j++;
              cnt = j - i;
            } else
              break;
          }
          if (i > 1 && !a[i - 1]) {
            for (int k = i; k < j; k += 2) {
              ans.push_back(k - 1);
              a[k] = a[k + 1] = 0;
            }
          } else if (j <= n && !a[j]) {
            for (int k = j - 2; k >= i; k -= 2) {
              ans.push_back(k);
              a[k] = a[k + 1] = 0;
            }
          } else
            ok = 0;
          i = j - 1;
        }
      if (!ok)
        puts("NO");
      else {
        puts("YES");
        printf("%d\n", (int)ans.size());
        if (ans.size()) {
          for (int i : ans) printf("%d ", i);
          puts("");
        }
      }
    }
  }
  