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 double pi = acos(-1);
  const long long mod = 998244353;
  const double eps = 1e-10;
  inline long long read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch > '9' || ch < '0') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return x * f;
  }
  int n, m;
  int a[200005], s[2], st[200005], tp, sum[200005];
  int main() {
    int t = read();
    while (t--) {
      n = read();
      s[0] = s[1] = 0;
      for (int i = 1; i <= n; i++) a[i] = read(), s[a[i]]++;
      if (!s[0]) {
        puts("NO");
        continue;
      }
      if (s[1] & 1) {
        puts("NO");
        continue;
      }
      if (!s[1]) {
        puts("YES");
        puts("0");
      } else {
        tp = 0;
        int w = 0;
        for (int i = 1; i <= n; i++) {
          sum[i] = sum[i - 1] + a[i];
          if (a[i] == 0 && sum[i - 1] % 2 == 0 && (s[1] - sum[i - 1]) % 2 == 0) {
            w = i;
            break;
          }
        }
        if (!w) {
          int len = 0, LL = 0;
          for (int i = 1; i <= n && !w; i++) {
            if (a[i] == 0)
              len++;
            else {
              if (len & 1) w = i - 1, LL = len;
              len = 0;
            }
          }
          if (!w)
            puts("NO");
          else {
            for (int i = w - LL; i <= w - 3; i += 2) {
              st[++tp] = i;
              a[i + 1] = a[i + 2] = 1;
            }
            st[++tp] = w - 1;
            a[w - 1] = a[w + 1] = 0;
            puts("YES");
            for (int i = w; i <= n - 2; i++) {
              if (a[i] == 0) {
                if (a[i + 1] == 0) continue;
                if (a[i + 2] == 1) {
                  st[++tp] = i;
                  a[i + 1] = a[i + 2] = 0;
                  continue;
                }
                if (a[i + 3] == 1) {
                  st[++tp] = i + 1;
                  a[i + 1] = a[i + 3] = 0;
                  continue;
                }
                st[++tp] = i + 1;
                st[++tp] = i;
                a[i + 1] = a[i + 2] = 0, a[i + 3] = 1;
              }
            }
            for (int i = w; i >= 3; i--) {
              if (a[i] == 0) {
                if (a[i - 1] == 0) continue;
                if (a[i - 2] == 1) {
                  st[++tp] = i - 2;
                  a[i - 1] = a[i - 2] = 0;
                  continue;
                }
                if (a[i - 3] == 1) {
                  st[++tp] = i - 3;
                  a[i - 1] = a[i - 3] = 0;
                  continue;
                }
                st[++tp] = i - 3;
                st[++tp] = i - 2;
                a[i - 1] = a[i - 2] = 0, a[i - 3] = 1;
              }
            }
            printf("%d\n", tp);
            for (int i = 1; i <= tp; i++) printf("%d ", st[i]);
            puts("");
          }
        } else {
          puts("YES");
          for (int i = w; i <= n - 2; i++) {
            if (a[i] == 0) {
              if (a[i + 1] == 0) continue;
              if (a[i + 2] == 1) {
                st[++tp] = i;
                a[i + 1] = a[i + 2] = 0;
                continue;
              }
              if (a[i + 3] == 1) {
                st[++tp] = i + 1;
                a[i + 1] = a[i + 3] = 0;
                continue;
              }
              st[++tp] = i + 1;
              st[++tp] = i;
              a[i + 1] = a[i + 2] = 0, a[i + 3] = 1;
            }
          }
          for (int i = w; i >= 3; i--) {
            if (a[i] == 0) {
              if (a[i - 1] == 0) continue;
              if (a[i - 2] == 1) {
                st[++tp] = i - 2;
                a[i - 1] = a[i - 2] = 0;
                continue;
              }
              if (a[i - 3] == 1) {
                st[++tp] = i - 3;
                a[i - 1] = a[i - 3] = 0;
                continue;
              }
              st[++tp] = i - 3;
              st[++tp] = i - 2;
              a[i - 1] = a[i - 2] = 0, a[i - 3] = 1;
            }
          }
          printf("%d\n", tp);
          for (int i = 1; i <= tp; i++) printf("%d ", st[i]);
          puts("");
        }
      }
    }
    return 0;
  }
  