#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
template <class T>
bool uin(T& a, const T& b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
bool uax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void solve() {
  int n;
  cin >> n;
  int cur = 0;
  vector<int> a(n);
  for (auto& x : a) {
    cin >> x;
    cur ^= x;
  }
  if (cur != 0) return void(cout << "NO\n");
  vector<int> ops;
  auto do_op = [&](int i) {
    assert(0 <= i && i < n - 2);
    int x = a[i] ^ a[i + 1] ^ a[i + 2];
    a[i] = a[i + 1] = a[i + 2] = x;
    ops.push_back(i);
  };
  auto solve = [&](int l, int r) {
    assert((r - l) % 2);
    for (int i = l; i < r - 2; i += 2) do_op(i);
    for (int i = r - 5; i >= l; i -= 2) do_op(i);
  };
  if (n & 1)
    solve(0, n);
  else {
    int j = 1;
    cur = a[0];
    for (; j + 1 < n; j += 2) {
      if (cur == 0) break;
      cur ^= a[j] ^ a[j + 1];
    }
    if (cur != 0) return void(cout << "NO\n");
    solve(0, j);
    solve(j, n);
  }
  for (auto x : a) assert(x == 0);
  cout << "YES\n";
  cout << int((ops).size()) << '\n';
  for (auto x : ops) cout << x + 1 << ' ';
  cout << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
template <class T>
bool uin(T& a, const T& b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
bool uax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void solve() {
  int n;
  cin >> n;
  int cur = 0;
  vector<int> a(n);
  for (auto& x : a) {
    cin >> x;
    cur ^= x;
  }
  if (cur != 0) return void(cout << "NO\n");
  vector<int> ops;
  auto do_op = [&](int i) {
    assert(0 <= i && i < n - 2);
    int x = a[i] ^ a[i + 1] ^ a[i + 2];
    a[i] = a[i + 1] = a[i + 2] = x;
    ops.push_back(i);
  };
  auto solve = [&](int l, int r) {
    assert((r - l) % 2);
    for (int i = l; i < r - 2; i += 2) do_op(i);
    for (int i = r - 5; i >= l; i -= 2) do_op(i);
  };
  if (n & 1)
    solve(0, n);
  else {
    int j = 1;
    cur = a[0];
    for (; j + 1 < n; j += 2) {
      if (cur == 0) break;
      cur ^= a[j] ^ a[j + 1];
    }
    if (cur != 0) return void(cout << "NO\n");
    solve(0, j);
    solve(j, n);
  }
  for (auto x : a) assert(x == 0);
  cout << "YES\n";
  cout << int((ops).size()) << '\n';
  for (auto x : ops) cout << x + 1 << ' ';
  cout << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) solve();
}

```