#include <bits/stdc++.h>
using namespace std;
int a[200005];
int p[200005];
int n, m;
pair<int, int> t[200005];
int len;
vector<int> ans;
void solve(int x) {
  ans.emplace_back(x);
  int val = a[x] ^ a[x + 1] ^ a[x + 2];
  a[x] = a[x + 1] = a[x + 2] = val;
}
void print() {
  cout << "YES\n";
  cout << ans.size() << '\n';
  if (ans.empty()) return;
  for (auto u : ans) cout << u << ' ';
  cout << '\n';
}
void solve() {
  cin >> n;
  ans.clear();
  m = len = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++)
    if (a[i]) p[++m] = i;
  if ((m & 1) || n == m) {
    cout << "NO\n";
    return;
  }
  int l = 1;
  p[m + 1] = 1e9;
  for (int i = 2; i <= m + 1; i++) {
    if (p[i] > p[i - 1] + 1) {
      t[++len] = make_pair(p[l], p[i - 1]);
      l = i;
    }
  }
  for (int i = 1; i <= len; i++)
    if ((t[i].second - t[i].first + 1) & 1) {
      int pos = t[i].second;
      while (t[i + 1].first - pos + 1 > 3) {
        solve(pos);
        pos += 2;
      }
      if (pos == t[i + 1].first - 1) {
        t[i + 1].first = t[i].first;
        continue;
      }
      while (pos >= t[i].first) {
        solve(pos);
        pos -= 2;
      }
      t[i + 1].first++;
    } else {
      if (t[i].first != 1) {
        for (int j = t[i].first; j <= t[i].second; j += 2) {
          solve(j - 1);
        }
      } else if (t[i].second != n) {
        for (int j = t[i].second; j >= t[i].first; j -= 2) {
          solve(j - 1);
        }
      }
    }
  for (int i = 1; i <= n; i++)
    if (a[i]) {
      cout << "NO\n";
      return;
    }
  print();
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  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;
int a[200005];
int p[200005];
int n, m;
pair<int, int> t[200005];
int len;
vector<int> ans;
void solve(int x) {
  ans.emplace_back(x);
  int val = a[x] ^ a[x + 1] ^ a[x + 2];
  a[x] = a[x + 1] = a[x + 2] = val;
}
void print() {
  cout << "YES\n";
  cout << ans.size() << '\n';
  if (ans.empty()) return;
  for (auto u : ans) cout << u << ' ';
  cout << '\n';
}
void solve() {
  cin >> n;
  ans.clear();
  m = len = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++)
    if (a[i]) p[++m] = i;
  if ((m & 1) || n == m) {
    cout << "NO\n";
    return;
  }
  int l = 1;
  p[m + 1] = 1e9;
  for (int i = 2; i <= m + 1; i++) {
    if (p[i] > p[i - 1] + 1) {
      t[++len] = make_pair(p[l], p[i - 1]);
      l = i;
    }
  }
  for (int i = 1; i <= len; i++)
    if ((t[i].second - t[i].first + 1) & 1) {
      int pos = t[i].second;
      while (t[i + 1].first - pos + 1 > 3) {
        solve(pos);
        pos += 2;
      }
      if (pos == t[i + 1].first - 1) {
        t[i + 1].first = t[i].first;
        continue;
      }
      while (pos >= t[i].first) {
        solve(pos);
        pos -= 2;
      }
      t[i + 1].first++;
    } else {
      if (t[i].first != 1) {
        for (int j = t[i].first; j <= t[i].second; j += 2) {
          solve(j - 1);
        }
      } else if (t[i].second != n) {
        for (int j = t[i].second; j >= t[i].first; j -= 2) {
          solve(j - 1);
        }
      }
    }
  for (int i = 1; i <= n; i++)
    if (a[i]) {
      cout << "NO\n";
      return;
    }
  print();
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

```