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 read() {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= ch == '-', ch = getchar();
    while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
    return f ? -x : x;
  }
  template <typename T>
  void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x / 10);
    putchar(x % 10 + '0');
  }
  template <typename T>
  void print(T x, char let) {
    print(x), putchar(let);
  }
  const int N = 2e5 + 5;
  int t, n, a[N], c[N], tot, flag;
  void solve() {
    n = read();
    tot = 0;
    flag = 1;
    for (int i = 1; i <= n; i++) a[i] = read();
    int i;
    if (a[1] == 0) {
      i = 2;
    } else {
      int j = 1;
      while ((a[j] == 1 && (a[j] + a[j + 1] + a[j + 2]) % 2 == 1) ||
             (a[j] == 0 && (a[j] + a[j + 1] + a[j + 2]) % 2 == 0)) {
        j += 2;
        if (j == n - 1) {
          flag = 0;
          break;
        }
        if (j == n) {
          flag = 0;
          break;
        }
      }
      if (flag == 0) {
        cout << "NO" << endl;
        return;
      } else {
        for (int k = j; k >= 1; k -= 2) c[++tot] = k;
        for (int k = 1; k <= j; k += 2)
          if (a[k] == 0) c[++tot] = k;
      }
      i = j + 3;
    }
    while (i <= n) {
      if (a[i] == 1) {
        if (i == n - 1) {
          if (a[n] == 1) {
            c[++tot] = i - 1;
            break;
          } else {
            flag = 0;
            break;
          }
        }
        if (i == n) {
          flag = 0;
          break;
        }
        if (a[i + 1] + a[i + 2] == 1) {
          c[++tot] = i;
          i += 3;
        } else if (a[i + 1] == 1) {
          c[++tot] = i - 1;
          i += 2;
        } else {
          c[++tot] = i;
          c[++tot] = i - 1;
          a[i + 2] = 1;
          i += 2;
        }
      } else {
        i++;
      }
    }
    if (flag == 0) {
      cout << "NO" << endl;
      return;
    } else {
      cout << "YES" << endl;
      cout << tot << endl;
      if (tot > 0) {
        for (int i = 1; i <= tot; i++) cout << c[i] << " ";
        cout << endl;
      }
    }
  }
  int main() {
    t = read();
    while (t--) {
      solve();
    }
    return 0;
  }
  