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>
  #pragma GCC optimize("Ofast,no-stack-protector")
  using namespace std;
  const int maxn = (int)2e5 + 10;
  int v[maxn];
  int n;
  vector<int> ans;
  void make(int pos) {
    assert(pos <= n - 3);
    int x = v[pos] ^ v[pos + 1] ^ v[pos + 2];
    v[pos] = v[pos + 1] = v[pos + 2] = x;
  }
  void solve(int l, int r) {
    if (r - l + 1 < 3) {
      assert(v[l] == 0 && v[r] == 0);
      return;
    }
    if (v[l] == 0) {
      if (v[l + 1] == 0) {
      } else if (v[l + 2] == 1) {
        ans.push_back(l);
        make(l);
      } else {
        ans.push_back(l + 1);
        make(l + 1);
        if (v[l + 1] == 1) {
          ans.push_back(l);
          make(l);
        }
      }
      solve(l + 1, r);
    } else {
      assert(v[r] == 0);
      if (v[r - 1] == 0) {
      } else if (v[r - 2] == 1) {
        ans.push_back(r - 2);
        make(r - 2);
      } else {
        ans.push_back(r - 3);
        make(r - 3);
        if (v[r - 1] == 1) {
          ans.push_back(r - 2);
          make(r - 2);
        }
      }
      solve(l, r - 1);
    }
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      ans.clear();
      scanf("%d", &n);
      int s = 0;
      for (int i = 0; i < n; i++) {
        scanf("%d", &v[i]);
        s ^= v[i];
      }
      if (s) {
        printf("NO\n");
        continue;
      }
      int num = 0;
      bool ok = false;
      for (int i = 0; i < n; i++) {
        if (v[i] == 1) {
          num ^= 1;
        } else {
          if (num == 0) {
            solve(0, i);
            solve(i, n - 1);
            ok = true;
            break;
          }
        }
      }
      if (!ok) {
        int last = 0;
        for (int i = 1; i < n; i++) {
          if (v[i] == 1) {
            if ((i - last) % 2 == 0) {
              for (int j = last; j + 2 < i - 1; j += 2) {
                ans.push_back(j);
                make(j);
              }
              for (int j = i; j - 2 >= last; j -= 2) {
                ans.push_back(j - 2);
                make(j - 2);
              }
              solve(0, last);
              solve(i, n - 1);
              ok = true;
              break;
            }
            last = i;
          }
        }
      }
      if (!ok) {
        printf("NO\n");
        continue;
      }
      printf("YES\n");
      printf("%d\n", (int)ans.size());
      assert((int)ans.size() <= n);
      for (int x : ans) {
        printf("%d ", x + 1);
      }
      for (int i = 0; i < n; i++) {
        assert(v[i] == 0);
      }
      printf("\n");
    }
  }
  