#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int N, A[maxn];
vector<int> ans;
bool rev;
int f(int i) { return (A[i] + A[i + 1] + A[i + 2]) & 1; }
void op(int i) {
  if (i + 2 > N) return;
  if (!rev)
    ans.emplace_back(i);
  else
    ans.emplace_back(N - i + 1 - 2);
  A[i] = A[i + 1] = A[i + 2] = f(i);
}
void go(int i) {
  if (i > N) return;
  if (!A[i]) {
    go(i + 1);
    return;
  }
  int j = i;
  while (j <= N && A[j] == 1) ++j;
  if (j > N) return;
  if ((j - i) % 2 == 0) {
    for (int k = j - 2; k >= 1 && A[k] && A[k + 1]; k -= 2) op(k);
    go(j);
    return;
  }
  if (j + 1 <= N && A[j + 1] == 1) {
    op(j - 1);
    for (int k = j - 3; k >= 1 && A[k] && A[k + 1]; k -= 2) op(k);
    go(j + 2);
    return;
  }
  op(j - 1);
  go(j + 1);
}
void solve() {
  cin >> N;
  A[N + 1] = A[N + 2] = 1;
  for (int i = 1; i <= N; ++i) cin >> A[i];
  if (count(A + 1, A + 1 + N, 1) & 1) {
    cout << "NO" << '\n';
    return;
  }
  ans.clear();
  rev = false;
  go(1);
  reverse(A + 1, A + 1 + N);
  rev = true;
  go(1);
  reverse(A + 1, A + 1 + N);
  rev = false;
  go(1);
  for (int i = 1; i <= N; ++i)
    if (A[i]) {
      cout << "NO" << '\n';
      return;
    }
  for (auto& x : ans)
    if (x + 2 > N) {
      cout << "NO" << '\n';
      return;
    }
  cout << "YES" << '\n';
  cout << ans.size() << '\n';
  copy((ans).begin(), (ans).end(), ostream_iterator<int>(cout, " ")),
      cout << '\n';
}
int main(void) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int TC;
  cin >> TC;
  while (TC--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
const int maxn = 2e5 + 10;
int N, A[maxn];
vector<int> ans;
bool rev;
int f(int i) { return (A[i] + A[i + 1] + A[i + 2]) & 1; }
void op(int i) {
  if (i + 2 > N) return;
  if (!rev)
    ans.emplace_back(i);
  else
    ans.emplace_back(N - i + 1 - 2);
  A[i] = A[i + 1] = A[i + 2] = f(i);
}
void go(int i) {
  if (i > N) return;
  if (!A[i]) {
    go(i + 1);
    return;
  }
  int j = i;
  while (j <= N && A[j] == 1) ++j;
  if (j > N) return;
  if ((j - i) % 2 == 0) {
    for (int k = j - 2; k >= 1 && A[k] && A[k + 1]; k -= 2) op(k);
    go(j);
    return;
  }
  if (j + 1 <= N && A[j + 1] == 1) {
    op(j - 1);
    for (int k = j - 3; k >= 1 && A[k] && A[k + 1]; k -= 2) op(k);
    go(j + 2);
    return;
  }
  op(j - 1);
  go(j + 1);
}
void solve() {
  cin >> N;
  A[N + 1] = A[N + 2] = 1;
  for (int i = 1; i <= N; ++i) cin >> A[i];
  if (count(A + 1, A + 1 + N, 1) & 1) {
    cout << "NO" << '\n';
    return;
  }
  ans.clear();
  rev = false;
  go(1);
  reverse(A + 1, A + 1 + N);
  rev = true;
  go(1);
  reverse(A + 1, A + 1 + N);
  rev = false;
  go(1);
  for (int i = 1; i <= N; ++i)
    if (A[i]) {
      cout << "NO" << '\n';
      return;
    }
  for (auto& x : ans)
    if (x + 2 > N) {
      cout << "NO" << '\n';
      return;
    }
  cout << "YES" << '\n';
  cout << ans.size() << '\n';
  copy((ans).begin(), (ans).end(), ostream_iterator<int>(cout, " ")),
      cout << '\n';
}
int main(void) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int TC;
  cin >> TC;
  while (TC--) {
    solve();
  }
  return 0;
}

```