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;
  using P = pair<int, int>;
  int main() {
    cin.tie(0);
    ios_base::sync_with_stdio(0);
    int n, t, k, x;
    int arr[200200];
    int ans[200200], top;
    cin >> t;
    while (t--) {
      cin >> n;
      x = 0;
      for (int i = 0; i < n; i++) {
        cin >> arr[i];
        x += arr[i];
      }
      if (x % 2 == 1) {
        cout << "NO\n";
        continue;
      }
      top = 0;
      if (arr[0] == 1 && arr[n - 1] == 1) {
        for (k = 0; arr[k] == 1 && k < n; k++)
          ;
        if (k % 2 == 0) {
          if (k == n) {
            cout << "NO\n";
            continue;
          }
          for (int i = k - 1; i > 0; i -= 2) {
            ans[top++] = i;
            arr[i - 1] = 0;
            arr[i] = 0;
          }
        } else if (k + 1 < n && arr[k + 1] == 1) {
          ans[top++] = k;
          arr[k - 1] = 0;
          arr[k + 1] = 0;
          for (int i = k - 2; i > 0; i -= 2) {
            ans[top++] = i;
            arr[i - 1] = 0;
            arr[i] = 0;
          }
        } else {
          for (int i = k; i < n - 1; i += 2) {
            ans[top++] = i;
            x = arr[i - 1] ^ arr[i] ^ arr[i + 1];
            arr[i - 1] = x;
            arr[i] = x;
            arr[i + 1] = x;
            if (x == 0) {
              for (int j = i - 2; j > 0; j -= 2) {
                ans[top++] = j;
                arr[j] = 0;
                arr[j - 1] = 0;
              }
              break;
            }
          }
        }
      }
      if (arr[0] == 0) {
        for (int i = 1; i < n - 1; i++) {
          if (arr[i] == 1) {
            if (arr[i + 1] == 1) {
              ans[top++] = i;
              arr[i] = 0;
              arr[i + 1] = 0;
            } else if (arr[i + 2] == 1) {
              ans[top++] = i + 1;
              arr[i] = 0;
              arr[i + 2] = 0;
            } else {
              ans[top++] = i + 1;
              ans[top++] = i;
              arr[i] = 0;
              arr[i + 2] = 1;
              i++;
            }
          }
        }
      } else if (arr[n - 1] == 0) {
        for (int i = n - 2; i > 0; i--) {
          if (arr[i] == 1) {
            if (arr[i - 1] == 1) {
              ans[top++] = i;
              arr[i] = 0;
              arr[i - 1] = 0;
            } else if (arr[i - 2] == 1) {
              ans[top++] = i - 1;
              arr[i] = 0;
              arr[i - 2] = 0;
            } else {
              ans[top++] = i - 1;
              ans[top++] = i;
              arr[i] = 0;
              arr[i - 2] = 1;
              i--;
            }
          }
        }
      } else {
        cout << "NO\n";
        continue;
      }
      cout << "YES\n";
      cout << top << '\n';
      for (int i = 0; i < top; i++) {
        cout << ans[i] << ' ';
      }
      cout << '\n';
    }
    cout << flush;
    return 0;
  }
  