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 = 2e5 + 5;
  vector<int> ans;
  int a[N];
  void solve() {
    int n, sum = 0, i, j;
    scanf("%d", &n);
    ans.clear();
    for (i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
      sum = sum + a[i];
    }
    if (sum % 2 == 1) {
      printf("NO\n");
      return;
    }
    sum = 0;
    for (i = 1; i <= n; ++i) {
      if (a[i] == 1) {
        sum++;
      } else if (a[i] == 0) {
        if (sum % 2 == 0) {
          for (j = 2; j <= sum; j = j + 2) {
            ans.push_back(i - j);
            a[i - j] = 0;
            a[i - j + 1] = 0;
          }
          sum = 0;
        } else {
          if (a[i + 1] == 0) {
            ans.push_back(i - 1);
            a[i + 1] = 1;
            a[i] = 1;
            sum++;
          } else if (a[i + 1] == 1) {
            ans.push_back(i - 1);
            sum--;
            a[i - 1] = 0;
            a[i + 1] = 0;
            for (j = 2; j <= sum; j = j + 2) {
              ans.push_back(i - 1 - j);
              a[i - 1 - j] = 0;
              a[i - j] = 0;
            }
            sum = 0;
          }
        }
      }
    }
    int flag = 0;
    for (i = 1; i <= n; ++i) {
      if (a[i] == 0) {
        flag = 1;
        break;
      }
    }
    if (flag == 0) {
      printf("NO\n");
      return;
    }
    for (i = 1; i <= n; ++i) {
      if (a[i] == 1) {
        ans.push_back(i - 1);
        a[i] = a[i + 1] = 0;
      }
    }
    printf("YES\n");
    printf("%d\n", ans.size());
    for (i = 0; i < ans.size(); ++i) {
      printf("%d ", ans[i]);
      if (i == ans.size() - 1) cout << endl;
    }
    return;
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      solve();
    }
  }
  