#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n, a[N], cnt;
void solve() {
  bool flag = false;
  cnt = 0;
  cin >> n;
  vector<int> ones, ans;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i]) {
      ones.push_back(i);
      cnt++;
    }
  }
  if (cnt & 1) {
    cout << "NO" << endl;
    return;
  }
  for (int i = 1; i < cnt; i += 2) {
    int firstIndex = ones[i - 1], secondIndex = ones[i];
    vector<int> curr;
    for (int j = firstIndex; j < secondIndex; j += 2) {
      if (j + 2 <= secondIndex) {
        int x = a[j] ^ a[j + 1] ^ a[j + 2];
        a[j] = x, a[j + 1] = x, a[j + 2] = x;
        curr.push_back(j);
        ans.push_back(j);
      }
    }
    if ((secondIndex - firstIndex - 1) & 1) {
      flag = true;
      for (int j = (int)curr.size() - 2; j >= 0; j--) {
        ans.push_back(curr[j]);
        int x = a[curr[j]] ^ a[curr[j] + 1] ^ a[curr[j] + 2];
        a[curr[j]] = x, a[curr[j] + 1] = x, a[curr[j] + 2] = x;
      }
    }
  }
  if (!flag) {
    for (int i = 1; i <= n; i++) {
      if (!a[i]) {
        flag = true;
        break;
      }
    }
    if (!flag) {
      cout << "NO" << endl;
      return;
    }
  }
  for (int i = 1; i <= n; i++) {
    int firstIndex = i, secondIndex = i;
    if (!a[i]) continue;
    while (secondIndex + 1 <= n && a[secondIndex + 1] == 1) {
      secondIndex++;
    }
    if (firstIndex == secondIndex) {
      continue;
    }
    if (firstIndex == 1) {
      for (int j = secondIndex - 1; j >= firstIndex; j -= 2) {
        ans.push_back(j);
      }
    } else {
      for (int j = firstIndex - 1; j < secondIndex; j += 2) {
        ans.push_back(j);
      }
    }
    i = secondIndex;
  }
  cout << "YES" << endl;
  cout << ans.size() << endl;
  for (auto i : ans) {
    cout << i << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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 int N = 2e5 + 10;
int n, a[N], cnt;
void solve() {
  bool flag = false;
  cnt = 0;
  cin >> n;
  vector<int> ones, ans;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i]) {
      ones.push_back(i);
      cnt++;
    }
  }
  if (cnt & 1) {
    cout << "NO" << endl;
    return;
  }
  for (int i = 1; i < cnt; i += 2) {
    int firstIndex = ones[i - 1], secondIndex = ones[i];
    vector<int> curr;
    for (int j = firstIndex; j < secondIndex; j += 2) {
      if (j + 2 <= secondIndex) {
        int x = a[j] ^ a[j + 1] ^ a[j + 2];
        a[j] = x, a[j + 1] = x, a[j + 2] = x;
        curr.push_back(j);
        ans.push_back(j);
      }
    }
    if ((secondIndex - firstIndex - 1) & 1) {
      flag = true;
      for (int j = (int)curr.size() - 2; j >= 0; j--) {
        ans.push_back(curr[j]);
        int x = a[curr[j]] ^ a[curr[j] + 1] ^ a[curr[j] + 2];
        a[curr[j]] = x, a[curr[j] + 1] = x, a[curr[j] + 2] = x;
      }
    }
  }
  if (!flag) {
    for (int i = 1; i <= n; i++) {
      if (!a[i]) {
        flag = true;
        break;
      }
    }
    if (!flag) {
      cout << "NO" << endl;
      return;
    }
  }
  for (int i = 1; i <= n; i++) {
    int firstIndex = i, secondIndex = i;
    if (!a[i]) continue;
    while (secondIndex + 1 <= n && a[secondIndex + 1] == 1) {
      secondIndex++;
    }
    if (firstIndex == secondIndex) {
      continue;
    }
    if (firstIndex == 1) {
      for (int j = secondIndex - 1; j >= firstIndex; j -= 2) {
        ans.push_back(j);
      }
    } else {
      for (int j = firstIndex - 1; j < secondIndex; j += 2) {
        ans.push_back(j);
      }
    }
    i = secondIndex;
  }
  cout << "YES" << endl;
  cout << ans.size() << endl;
  for (auto i : ans) {
    cout << i << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

```