#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vl = vector<ll>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
void solve() {
  int n;
  cin >> n;
  vi a(n), freq(2);
  for (int& x : a) cin >> x, freq[x]++;
  if (freq[0] == 0) {
    cout << "NO\n";
    return;
  }
  vi ans;
  auto op = [&](int i) {
    ans.push_back(i);
    a[i] = a[i + 1] = a[i + 2] = a[i] ^ a[i + 1] ^ a[i + 2];
  };
  for (int z = 0; z < 2; z++) {
    int left = 0;
    for (int i = 0; i + 2 < n; i++) {
      if (a[i] + a[i + 1] + a[i + 2] == 2 && left % 2 == 0) {
        int idx = i;
        left += 2;
        while (left > 0) {
          op(idx);
          left -= 2;
          idx -= 2;
        }
        i++;
        continue;
      }
      if (i > 0 && a[i - 1] && a[i - 1] + a[i] + a[i + 1] == 1) op(i - 1);
      left += a[i];
    }
  }
  bool bad = ans.size() > n;
  for (int i : a)
    if (i) bad = 1;
  if (bad) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  cout << ans.size() << "\n";
  for (int i : ans) cout << i + 1 << " ";
  cout << "\n";
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vl = vector<ll>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
void solve() {
  int n;
  cin >> n;
  vi a(n), freq(2);
  for (int& x : a) cin >> x, freq[x]++;
  if (freq[0] == 0) {
    cout << "NO\n";
    return;
  }
  vi ans;
  auto op = [&](int i) {
    ans.push_back(i);
    a[i] = a[i + 1] = a[i + 2] = a[i] ^ a[i + 1] ^ a[i + 2];
  };
  for (int z = 0; z < 2; z++) {
    int left = 0;
    for (int i = 0; i + 2 < n; i++) {
      if (a[i] + a[i + 1] + a[i + 2] == 2 && left % 2 == 0) {
        int idx = i;
        left += 2;
        while (left > 0) {
          op(idx);
          left -= 2;
          idx -= 2;
        }
        i++;
        continue;
      }
      if (i > 0 && a[i - 1] && a[i - 1] + a[i] + a[i + 1] == 1) op(i - 1);
      left += a[i];
    }
  }
  bool bad = ans.size() > n;
  for (int i : a)
    if (i) bad = 1;
  if (bad) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  cout << ans.size() << "\n";
  for (int i : ans) cout << i + 1 << " ";
  cout << "\n";
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```