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>
  template <class T>
  inline T read() {
    T r = 0, f = 0;
    char c;
    while (!isdigit(c = getchar())) f |= (c == '-');
    while (isdigit(c)) r = (r << 1) + (r << 3) + (c ^ 48), c = getchar();
    return f ? -r : r;
  }
  template <class T>
  inline T min(T a, T b) {
    return a < b ? a : b;
  }
  template <class T>
  inline T max(T a, T b) {
    return a > b ? a : b;
  }
  template <class T>
  inline void swap(T &a, T &b) {
    T c = a;
    a = b;
    b = c;
  }
  inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
  inline long long qpow(long long a, int b) {
    long long ans = 1;
    for (; b; b >>= 1) {
      if (b & 1) (ans *= a) %= 1000000007;
      (a *= a) %= 1000000007;
    }
    return ans;
  }
  struct Z {
    int x;
    inline int val() const { return x; }
    inline int inv() const { return qpow(x, 1000000007 - 2); }
    Z(int x = 0) : x(x) {}
    Z operator-() const {
      return Z((1000000007 - x >= 1000000007 ? 1000000007 - x - 1000000007
                                             : 1000000007 - x));
    }
    Z &operator+=(const Z &z) {
      x = (x + z.x >= 1000000007 ? x + z.x - 1000000007 : x + z.x);
      return *this;
    }
    Z &operator-=(const Z &z) {
      x = (x - z.x < 0 ? x - z.x + 1000000007 : x - z.x);
      return *this;
    }
    Z &operator*=(const Z &z) {
      x = 1ll * x * z.x % 1000000007;
      return *this;
    }
    Z &operator/=(const Z &z) {
      x = 1ll * x * z.inv() % 1000000007;
      return *this;
    }
    Z operator+(const Z &z) const {
      return Z((x + z.x >= 1000000007 ? x + z.x - 1000000007 : x + z.x));
    }
    Z operator-(const Z &z) const {
      return Z((x - z.x < 0 ? x - z.x + 1000000007 : x - z.x));
    }
    Z operator*(const Z &z) const { return Z(1ll * x * z.x % 1000000007); }
    Z operator/(const Z &z) const { return Z(1ll * x * z.inv() % 1000000007); }
  };
  int n, a[202202];
  int s_ans, ans[202202];
  void solve(int add, int n) {
    for (int i = 1; i <= n - 2; i += 2) ans[++s_ans] = i + add;
    for (int i = n - 4; i > 0; i -= 2) ans[++s_ans] = i + add;
  }
  inline void out() {
    printf("YES\n%d\n", s_ans);
    for (int i = 1; i <= s_ans; i++) printf("%d ", ans[i]);
    puts("");
  }
  inline void work() {
    n = read<int>();
    int X = 0;
    bool ok = 0;
    for (int i = 1; i <= n; i++) X ^= a[i] = read<int>();
    if (X) return (void)puts("NO");
    s_ans = 0;
    if (n & 1) {
      solve(0, n);
      return out();
    }
    for (int i = 1, x = 0; i <= n; i++) {
      x ^= a[i];
      if ((i & 1) && !x) {
        solve(0, i);
        solve(i, n - i);
        return out();
      }
    }
    puts("NO");
  }
  int main() {
    int t = read<int>();
    while (t--) work();
    return 0;
  }
  