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>
  void sc(int &x) { scanf("%d", &x); }
  void sc(int &x, int &y) { scanf("%d%d", &x, &y); }
  void sc(int &x, int &y, int &z) { scanf("%d%d%d", &x, &y, &z); }
  void sc(long long &x) { scanf("%lld", &x); }
  void sc(long long &x, long long &y) { scanf("%lld%lld", &x, &y); }
  void sc(long long &x, long long &y, long long &z) {
    scanf("%lld%lld%lld", &x, &y, &z);
  }
  void sc(char *x) { scanf("%s", x); }
  void sc(char *x, char *y) { scanf("%s%s", x, y); }
  void sc(char *x, char *y, char *z) { scanf("%s%s%s", x, y, z); }
  void out(int x) { printf("%d\n", x); }
  void out(long long x) { printf("%lld\n", x); }
  void out(int x, int y) { printf("%d %d\n", x, y); }
  void out(long long x, long long y) { printf("%lld %lld\n", x, y); }
  void out(int x, int y, int z) { printf("%d %d %d\n", x, y, z); }
  void out(long long x, long long y, long long z) {
    printf("%lld %lld %lld\n", x, y, z);
  }
  void ast(long long x, long long l, long long r) { assert(x >= l && x <= r); }
  using namespace std;
  const int N = 3e5 + 5, mod = 998244353;
  int n, a[N];
  int sum(int x) { return a[x] + a[x + 1] + a[x + 2]; }
  void sol(int cas) {
    sc(n);
    for (int i = 1; i <= n; i++) sc(a[i]);
    vector<int> ans;
    int sm = 0, s = 0;
    for (int i = 1; i <= n; i++) sm += a[i];
    if (sm & 1) {
      printf("NO\n");
      return;
    }
    for (int i = 1; i <= n; i++)
      if (a[i] == 1) {
        int j = i;
        bool flag = false;
        while (j + 2 <= n) {
          if (sum(j) == 1) {
            ans.push_back(j);
            a[j + 1] = a[j + 2] = 1;
            j += 2;
          } else if (sum(j) == 2) {
            flag = true;
            for (int k = j; k >= i; k--)
              if (sum(k) == 2) {
                ans.push_back(k);
                a[k] = a[k + 1] = a[k + 2] = 0;
              }
            break;
          } else
            j += 2;
        }
        if (!flag) {
          if (i > 1) {
            for (int k = i - 1; k <= j - 1 && k <= n - 2; k++)
              if (sum(k) == 2) {
                ans.push_back(k);
                a[k] = a[k + 1] = a[k + 2] = 0;
              }
          }
        }
      }
    for (int i = 1; i <= n; i++)
      if (a[i]) {
        printf("NO\n");
        return;
      }
    printf("YES\n");
    printf("%d\n", ans.size());
    for (int i = 0; i < ans.size(); i++)
      printf(i == ans.size() - 1 ? "%d\n" : "%d ", ans[i]);
  }
  int main() {
    srand(time(0));
    int t = 1, cas = 0;
    scanf("%d", &t);
    while (t--) {
      sol(++cas);
    }
    return 0;
  }
  