#include <bits/stdc++.h>
using namespace std;
long long T, n, m, cnt, num[200100], ri[200100], qz[200100];
vector<long long> ans;
inline void work(long long u) {
  num[u] = num[u + 1] = num[u + 2] = (num[u] ^ num[u + 1] ^ num[u + 2]);
  ans.push_back(u);
}
inline void work(long long u, long long v) {
  long long i, j, k;
  i = u, j = v - 1;
  if ((j - i) & 1) {
    for (k = i; k < j; k += 2) work(k);
    for (k = j - 1; k > i; k -= 2) work(k - 2);
  } else {
    for (k = i; k < j; k += 2) work(k);
  }
}
int main() {
  long long i, j, p, q, t;
  cin >> T;
  while (T--) {
    cnt = 0;
    ans.clear();
    scanf("%lld", &n);
    for (i = 1; i <= n; i++) {
      scanf("%lld", &num[i]);
      qz[i] = qz[i - 1] + num[i];
      cnt += num[i];
    }
    if ((cnt & 1) || cnt == n) {
      puts("NO");
      continue;
    }
    ri[n + 1] = 0;
    for (i = n; i >= 1; i--) {
      if (!num[i])
        ri[i] = 0;
      else
        ri[i] = ri[i + 1] + 1;
    }
    for (p = 1; p < n; p++) {
      if (num[p] && !num[p + 1]) {
        if (!(qz[p] & 1)) continue;
        for (q = p + 1; q < n; q++)
          if (!num[q] && num[q + 1]) break;
        if (q < n)
          work(p, q + 1);
        else
          break;
      }
    }
    for (i = 1; i <= n - 2; i++)
      if (!num[i] && num[i + 1] && num[i + 2]) work(i);
    for (i = n - 2; i >= 1; i--)
      if (!num[i + 2] && num[i + 1] && num[i]) work(i);
    for (i = 1; i <= n; i++)
      if (num[i]) break;
    if (i <= n) {
      puts("NO");
      continue;
    }
    puts("YES");
    printf("%lld\n", (long long)ans.size());
    for (i = 0; i < ans.size(); i++) printf("%lld ", ans[i]);
    puts("");
  }
}

### Prompt

Generate a Cpp solution to the following problem:

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].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long T, n, m, cnt, num[200100], ri[200100], qz[200100];
vector<long long> ans;
inline void work(long long u) {
  num[u] = num[u + 1] = num[u + 2] = (num[u] ^ num[u + 1] ^ num[u + 2]);
  ans.push_back(u);
}
inline void work(long long u, long long v) {
  long long i, j, k;
  i = u, j = v - 1;
  if ((j - i) & 1) {
    for (k = i; k < j; k += 2) work(k);
    for (k = j - 1; k > i; k -= 2) work(k - 2);
  } else {
    for (k = i; k < j; k += 2) work(k);
  }
}
int main() {
  long long i, j, p, q, t;
  cin >> T;
  while (T--) {
    cnt = 0;
    ans.clear();
    scanf("%lld", &n);
    for (i = 1; i <= n; i++) {
      scanf("%lld", &num[i]);
      qz[i] = qz[i - 1] + num[i];
      cnt += num[i];
    }
    if ((cnt & 1) || cnt == n) {
      puts("NO");
      continue;
    }
    ri[n + 1] = 0;
    for (i = n; i >= 1; i--) {
      if (!num[i])
        ri[i] = 0;
      else
        ri[i] = ri[i + 1] + 1;
    }
    for (p = 1; p < n; p++) {
      if (num[p] && !num[p + 1]) {
        if (!(qz[p] & 1)) continue;
        for (q = p + 1; q < n; q++)
          if (!num[q] && num[q + 1]) break;
        if (q < n)
          work(p, q + 1);
        else
          break;
      }
    }
    for (i = 1; i <= n - 2; i++)
      if (!num[i] && num[i + 1] && num[i + 2]) work(i);
    for (i = n - 2; i >= 1; i--)
      if (!num[i + 2] && num[i + 1] && num[i]) work(i);
    for (i = 1; i <= n; i++)
      if (num[i]) break;
    if (i <= n) {
      puts("NO");
      continue;
    }
    puts("YES");
    printf("%lld\n", (long long)ans.size());
    for (i = 0; i < ans.size(); i++) printf("%lld ", ans[i]);
    puts("");
  }
}

```