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;
  int a[200001];
  void op(int i) {
    int res = a[i] ^ a[i + 1] ^ a[i + 2];
    a[i] = a[i + 1] = a[i + 2] = res;
  }
  vector<int> push_left(int i) {
    vector<int> v;
    while (i - 2 >= 1 && a[i] == 0 && a[i - 1] == 1 && a[i - 2] == 1) {
      op(i - 2);
      v.push_back(i - 2);
      i -= 2;
    }
    return v;
  }
  vector<int> push_right(int i, int n) {
    vector<int> v;
    while (i + 2 <= n && a[i] == 0 && a[i + 1] == 1 && a[i + 2] == 1) {
      op(i);
      v.push_back(i);
      i += 2;
    }
    return v;
  }
  int main() {
    int i, t;
    for (scanf("%d", &t); t--;) {
      int n;
      vector<int> v, ans;
      int cum = 0;
      for (scanf("%d", &n), i = 1; i <= n; ++i) {
        scanf("%d", a + i);
        cum += a[i];
      }
      if (cum == 0) {
        printf("YES\n0\n");
        continue;
      } else if ((cum & 1) || cum == n) {
        printf("NO\n");
        continue;
      }
      int cur = 1;
      if (a[1]) {
        cum = a[1];
        cur = 2;
        while (cur <= n && a[cur]) cum += a[cur], ++cur;
        if (cum & 1) {
          --cur;
          bool y = 1;
          while (cur <= n && a[cur]) {
            if (cur + 2 <= n) {
              if (a[cur] + a[cur + 1] + a[cur + 2] == 3) {
                cur += 2;
              } else if (a[cur + 1] + a[cur + 2] == 0) {
                ans.push_back(cur);
                op(cur);
                cur += 2;
              } else if (a[cur + 1] == 0 && a[cur + 2] == 1) {
                ans.push_back(cur);
                op(cur);
                auto push_vec = push_left(cur);
                ans.insert(ans.end(), push_vec.begin(), push_vec.end());
                cur += 3;
                break;
              } else {
                auto push_vec = push_left(cur + 2);
                ans.insert(ans.end(), push_vec.begin(), push_vec.end());
                cur += 3;
                break;
              }
            } else {
              y = 0;
              break;
            }
          }
          if (y == 0) {
            printf("NO\n");
            continue;
          }
        } else {
          ans = push_left(cur);
        }
      }
      cur = 1;
      while (cur + 2 <= n) {
        if (a[cur] == 0 && a[cur + 1] + a[cur + 2] == 2) {
          op(cur);
          ans.push_back(cur);
          cur += 2;
        } else if (a[cur]) {
          int st = cur - 1;
          while (1) {
            if (cur + 2 <= n && a[cur + 1] + a[cur + 2] == 2)
              cur += 2;
            else if (cur + 2 <= n && a[cur + 1] + a[cur + 2] == 0) {
              op(cur);
              ans.push_back(cur);
              cur += 2;
            } else if (cur + 2 <= n && a[cur + 1] == 0 && a[cur + 2] == 1) {
              op(cur);
              ans.push_back(cur);
              break;
            } else {
              break;
            }
          }
          auto push_vec = push_right(st, n);
          ans.insert(ans.end(), push_vec.begin(), push_vec.end());
        } else
          ++cur;
      }
      cum = 0;
      for (i = 1; i <= n; ++i) cum += a[i];
      if (cum == 0) {
        printf("YES\n%d\n", ans.size());
        for (auto x : ans) printf("%d ", x);
        puts("");
      } else
        printf("NO\n");
    }
    return 0;
  }
  