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;
  long long n, nn, a[200069], kd[200069], wg[200069], sq[200069], zs;
  void op(long long x) {
    long long i, sm = a[x] + a[x + 1] + a[x + 2];
    if (!sm) {
      return;
    }
    zs++;
    sq[zs] = x;
    for (i = x; i <= x + 2; i++) {
      a[i] = sm % 2;
    }
  }
  int main() {
    long long t, rr, i, j, c, e;
    scanf("%lld", &t);
    for (rr = 0; rr < t; rr++) {
      scanf("%lld", &n);
      c = 0;
      nn = 0;
      for (i = 1; i <= n; i++) {
        scanf("%lld", a + i);
        c += a[i];
        if (i == 1 || a[i] != a[i - 1]) {
          nn++;
          kd[nn] = a[i];
          wg[nn] = 0;
        }
        wg[nn]++;
      }
      if (c % 2 || c == n) {
        printf("NO\n");
        continue;
      }
      zs = 0;
      if (kd[1] && kd[nn]) {
        for (i = 2; i < nn; i++) {
          if (!kd[i] && wg[i] % 2) {
            break;
          }
        }
        if (i >= nn) {
          if (wg[1] % 2 == 0) {
            for (i = wg[1] - 1; i > 0; i -= 2) {
              op(i);
            }
          } else if (wg[nn] % 2 == 0) {
            for (i = n - wg[nn]; i <= n - 2; i += 2) {
              op(i);
            }
          } else {
            for (i = 2; i < nn; i++) {
              if (kd[i] && wg[i] % 2) {
                break;
              }
            }
            if (i >= nn) {
              printf("NO\n");
              continue;
            }
            e = i;
            c = 0;
            for (i = 1; i <= e; i++) {
              c += wg[i];
            }
            for (i = c - 1; i >= c - wg[e]; i -= 2) {
              op(i);
            }
          }
        }
      }
      c = 0;
      for (i = 1; i <= n - 2; i++) {
        c += a[i];
        if (a[i] && !a[i + 1] && !a[i + 2] && c % 2) {
          op(i);
        }
      }
      c = 0;
      for (i = 1; i <= n - 2; i++) {
        if ((a[i] + a[i + 1] + a[i + 2]) % 2 == 0 && c % 2 == 0) {
          op(i);
          for (j = i - 1; j; j--) {
            if ((a[j] + a[j + 1] + a[j + 2]) % 2 == 0) {
              op(j);
            }
          }
          for (j = i + 1; j <= n - 2; j++) {
            if ((a[j] + a[j + 1] + a[j + 2]) % 2 == 0) {
              op(j);
            }
          }
          break;
        }
        c += a[i];
      }
      printf("YES\n%lld\n", zs);
      for (i = 1; i <= zs; i++) {
        printf("%lld%c", sq[i], " \n"[i == zs]);
      }
    }
  }
  