#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MaxN = 500100;
int N;
vector<int> a;
bool pref[MaxN];
int breakpoint;
void check(vector<int> num, const vector<int> &ops) {
  for (auto mid : ops) {
    int r = num[mid - 1] ^ num[mid] ^ num[mid + 1];
    for (int z = mid - 1; z <= mid + 1; z++) num[z] = r;
  }
  for (int i = (1); i < (N + 1); i++) assert(num[i] == 0);
}
bool conditions() {
  if (pref[N] != 0) return false;
  for (int i = (1); i < (N + 1); i++) {
    if (i % 2 == 1 && pref[i] == 0) {
      breakpoint = i;
      return true;
    }
  }
  return false;
}
void solve() {
  if (!conditions()) {
    cout << "NO" << endl;
    return;
  };
  if (N % 2 == 1) breakpoint = N;
  assert(breakpoint % 2 == 1);
  vector<int> ops;
  for (int i = 2; i <= breakpoint - 1; i += 2) {
    ops.push_back(i);
  }
  for (int i = breakpoint - 3; i >= 2; i -= 2) {
    ops.push_back(i);
  }
  for (int i = breakpoint + 2; i <= N - 1; i += 2) {
    ops.push_back(i);
  }
  for (int i = N - 3; i >= breakpoint + 2; i -= 2) {
    ops.push_back(i);
  }
  cout << "YES" << endl;
  cout << ops.size() << endl;
  for (auto mid : ops) {
    cout << mid - 1 << " ";
  }
  cout << endl;
  check(a, ops);
}
void load() {
  breakpoint = -1;
  cin >> N;
  a.clear();
  a.resize(N + 1);
  a[0] = -100;
  for (int i = (0); i < (N); i++) cin >> a[i + 1];
  pref[0] = 0;
  for (int i = (0); i < (N); i++) pref[i + 1] = pref[i] ^ a[i + 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  int T;
  cin >> T;
  for (int tt = (0); tt < (T); tt++) {
    load();
    solve();
  }
  return 0;
}

### Prompt

Generate 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MaxN = 500100;
int N;
vector<int> a;
bool pref[MaxN];
int breakpoint;
void check(vector<int> num, const vector<int> &ops) {
  for (auto mid : ops) {
    int r = num[mid - 1] ^ num[mid] ^ num[mid + 1];
    for (int z = mid - 1; z <= mid + 1; z++) num[z] = r;
  }
  for (int i = (1); i < (N + 1); i++) assert(num[i] == 0);
}
bool conditions() {
  if (pref[N] != 0) return false;
  for (int i = (1); i < (N + 1); i++) {
    if (i % 2 == 1 && pref[i] == 0) {
      breakpoint = i;
      return true;
    }
  }
  return false;
}
void solve() {
  if (!conditions()) {
    cout << "NO" << endl;
    return;
  };
  if (N % 2 == 1) breakpoint = N;
  assert(breakpoint % 2 == 1);
  vector<int> ops;
  for (int i = 2; i <= breakpoint - 1; i += 2) {
    ops.push_back(i);
  }
  for (int i = breakpoint - 3; i >= 2; i -= 2) {
    ops.push_back(i);
  }
  for (int i = breakpoint + 2; i <= N - 1; i += 2) {
    ops.push_back(i);
  }
  for (int i = N - 3; i >= breakpoint + 2; i -= 2) {
    ops.push_back(i);
  }
  cout << "YES" << endl;
  cout << ops.size() << endl;
  for (auto mid : ops) {
    cout << mid - 1 << " ";
  }
  cout << endl;
  check(a, ops);
}
void load() {
  breakpoint = -1;
  cin >> N;
  a.clear();
  a.resize(N + 1);
  a[0] = -100;
  for (int i = (0); i < (N); i++) cin >> a[i + 1];
  pref[0] = 0;
  for (int i = (0); i < (N); i++) pref[i + 1] = pref[i] ^ a[i + 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  int T;
  cin >> T;
  for (int tt = (0); tt < (T); tt++) {
    load();
    solve();
  }
  return 0;
}

```