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;
  inline int read() {
    int x = 0, w = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      w |= ch == '-';
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = (x << 3) + (x << 1) + ch - '0';
      ch = getchar();
    }
    return w ? -x : x;
  }
  inline long long readll() {
    long long x = 0, w = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      w |= ch == '-';
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = (x << 3) + (x << 1) + ch - '0';
      ch = getchar();
    }
    return w ? -x : x;
  }
  int Tc;
  int n, a[200005], flag;
  int ans[200005], cnt;
  void work(int x) {
    ans[++cnt] = x;
    int tmp = a[x] ^ a[x + 1] ^ a[x + 2];
    a[x] = a[x + 1] = a[x + 2] = tmp;
  }
  void solve() {
    cnt = 0;
    scanf("%d", &n);
    int sum = 0;
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      sum += a[i];
    }
    if (sum & 1) {
      printf("NO\n");
      return;
    }
    sum = 0, flag = 0;
    for (int i = 1; i <= n; i++) {
      sum ^= a[i];
      if (sum == 0 && a[i] == 0) flag = i;
    }
    if (!flag) {
      int last = 1;
      for (int i = 1; i <= n; i++) {
        if (a[i] && ((i - last) & 1)) flag = i - 1;
        if (a[i]) last = i + 1;
      }
      if (!flag) {
        printf("NO\n");
        return;
      }
      for (; !a[flag - 1]; flag -= 2) work(flag - 1);
      work(flag - 1);
      flag--;
    }
    for (int i = flag - 1; i >= 1; i--) {
      if (!a[i]) continue;
      if (a[i - 1])
        work(i - 1);
      else {
        work(i - 2);
        if (a[i] == 1) work(i - 1);
      }
    }
    for (int i = flag + 1; i <= n; i++) {
      if (!a[i]) continue;
      if (a[i + 1])
        work(i - 1);
      else {
        work(i);
        if (a[i] == 1) work(i - 1);
      }
    }
    printf("YES\n%d\n", cnt);
    for (int i = 1; i <= cnt; i++) printf("%d ", ans[i]);
    printf("\n");
  }
  int main() {
    Tc = read();
    while (Tc--) solve();
    return 0;
  }
  