#include <bits/stdc++.h>
using namespace std;
const long long Mod = 1000000007LL;
const int N = 2e5 + 10;
const long long Inf = 2242545357980376863LL;
const long long Log = 30;
int a[N];
int n;
vector<int> V;
void Do(int i) {
  int x = a[i] ^ a[i + 1] ^ a[i + 2];
  a[i] = a[i + 1] = a[i + 2] = x;
  V.push_back(i);
}
int Main() {
  cin >> n;
  V.clear();
  int sm = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sm += a[i];
  }
  if (sm % 2 == 1) return cout << "NO\n", 0;
  if (a[1] == 0) {
  } else {
    int idx = -1;
    for (int i = 2; i + 1 <= n; i += 2) {
      if (a[i] != a[i + 1]) {
        idx = i;
        break;
      }
    }
    if (idx == -1) return cout << "NO\n", 0;
    for (int i = idx - 1; i >= 1; i -= 2) Do(i);
  }
  for (int i = 1; i <= n; i++) {
    if (i == n) break;
    if (a[i + 1] == 0) continue;
    if (a[i + 2] == 1) {
      Do(i);
      continue;
    }
    assert(i + 3 <= n);
    Do(i + 1);
    Do(i);
  }
  cout << "YES\n" << V.size() << '\n';
  for (auto x : V) cout << x << ' ';
  cout << '\n';
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) Main();
  return 0;
}

### Prompt

Your task is to create 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;
const long long Mod = 1000000007LL;
const int N = 2e5 + 10;
const long long Inf = 2242545357980376863LL;
const long long Log = 30;
int a[N];
int n;
vector<int> V;
void Do(int i) {
  int x = a[i] ^ a[i + 1] ^ a[i + 2];
  a[i] = a[i + 1] = a[i + 2] = x;
  V.push_back(i);
}
int Main() {
  cin >> n;
  V.clear();
  int sm = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sm += a[i];
  }
  if (sm % 2 == 1) return cout << "NO\n", 0;
  if (a[1] == 0) {
  } else {
    int idx = -1;
    for (int i = 2; i + 1 <= n; i += 2) {
      if (a[i] != a[i + 1]) {
        idx = i;
        break;
      }
    }
    if (idx == -1) return cout << "NO\n", 0;
    for (int i = idx - 1; i >= 1; i -= 2) Do(i);
  }
  for (int i = 1; i <= n; i++) {
    if (i == n) break;
    if (a[i + 1] == 0) continue;
    if (a[i + 2] == 1) {
      Do(i);
      continue;
    }
    assert(i + 3 <= n);
    Do(i + 1);
    Do(i);
  }
  cout << "YES\n" << V.size() << '\n';
  for (auto x : V) cout << x << ' ';
  cout << '\n';
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) Main();
  return 0;
}

```