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;
  const int N = 222222;
  int n, a[N];
  vector<int> v, s;
  void solver(int x, int y) {
    while (y - x >= 3) {
      s.push_back(y - 2);
      s.push_back(y - 1);
      y -= 2;
    }
    s.push_back(x);
  }
  void solvel(int x, int y) {
    while (y - x >= 3) {
      s.push_back(x);
      s.push_back(x - 1);
      x += 2;
    }
    s.push_back(y - 2);
  }
  void solvem(int x, int y) {
    int i;
    for (i = x; i < y; i += 2) s.push_back(i);
    for (i = y - 4; i >= x; i -= 2) s.push_back(i);
  }
  int main() {
    int T, i, x;
    scanf("%d", &T);
    while (T--) {
      scanf("%d", &n);
      x = 0;
      s.clear();
      v.clear();
      for (i = 1; i <= n; i++) {
        scanf("%d", a + i);
        x ^= a[i];
        if (a[i]) v.push_back(i);
      }
      if (x) {
        printf("NO\n");
        continue;
      }
      if (!a[n]) {
        for (i = v.size() - 1; i >= 0; i -= 2) solver(v[i - 1], v[i]);
        printf("YES\n");
        printf("%d\n", s.size());
        for (i = 0; i < s.size(); i++) printf("%d ", s[i]);
        printf("\n");
        continue;
      }
      if (!a[1]) {
        for (i = 0; i < v.size(); i += 2) solvel(v[i], v[i + 1]);
        printf("YES\n");
        printf("%d\n", s.size());
        for (i = 0; i < s.size(); i++) printf("%d ", s[i]);
        printf("\n");
        continue;
      }
      x = -1;
      for (i = 1; i < v.size() - 1; i += 2)
        if (v[i] < v[i + 1] - 1) x = i;
      if (x >= 0) {
        for (i = x; i >= 0; i -= 2) solver(v[i - 1], v[i]);
        for (i = x + 1; i < v.size(); i += 2) solvel(v[i], v[i + 1]);
        printf("YES\n");
        printf("%d\n", s.size());
        for (i = 0; i < s.size(); i++) printf("%d ", s[i]);
        printf("\n");
        continue;
      }
      x = -1;
      for (i = 0; i < v.size() - 1; i++)
        if (v[i] % 2 == v[i + 1] % 2) x = i;
      if (x >= 0) {
        solvem(v[x], v[x + 1]);
        for (i = x - 1; i >= 0; i -= 2) solver(v[i - 1], v[i]);
        for (i = x + 2; i < v.size(); i += 2) solvel(v[i], v[i + 1]);
        printf("YES\n");
        printf("%d\n", s.size());
        for (i = 0; i < s.size(); i++) printf("%d ", s[i]);
        printf("\n");
        continue;
      }
      printf("NO\n");
    }
    return 0;
  }
  