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].
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int T, n;
int a[N], ans[N], tot = 0;
int ask(int first) { return a[first] ^ a[first + 1] ^ a[first + 2]; }
void flip(int first) {
  if (a[first] || a[first + 1] || a[first + 2]) {
    a[first] = a[first + 1] = a[first + 2] =
        a[first] ^ a[first + 1] ^ a[first + 2];
    ans[++tot] = first;
  }
  return;
}
void solve() {
  tot = 0;
  cin >> n;
  int parity = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    parity ^= a[i];
  }
  if (parity) return void(cout << "NO" << endl);
  parity = a[1] ^ a[2];
  for (int i = 1; i <= n - 2; i++) {
    parity ^= a[i + 2];
    if (ask(i) == 0 && !parity) {
      flip(i);
      break;
    }
  }
  parity = 0;
  int pos = -1;
  parity = a[1] ^ a[2] ^ a[3];
  for (int i = 4; i <= n; i++) {
    parity ^= a[i];
    if (a[i] == 0 && !parity) {
      pos = i;
      break;
    }
  }
  for (int i = pos - 1; i >= 3; i--) {
    if (a[i] == 0) continue;
    if (ask(i - 2) == 0)
      flip(i - 2);
    else
      flip(i - 2), flip(i - 1);
  }
  flip(1);
  pos = n;
  parity = a[n] ^ a[n - 1] ^ a[n - 2];
  for (int i = n - 3; i >= 1; i--) {
    parity ^= a[i];
    if (a[i] == 0 && !parity) {
      pos = i;
      break;
    }
  }
  for (int i = pos + 1; i <= n - 2; i++) {
    if (a[i] == 0) continue;
    if (ask(i) == 0) {
      flip(i);
    } else {
      flip(i), flip(i - 1);
    }
  }
  flip(n - 2);
  for (int i = 1; i <= n; i++) {
    if (a[i]) {
      cout << "NO" << endl;
      return;
    }
  }
  cout << "YES" << endl;
  cout << tot << endl;
  for (int i = 1; i <= tot; i++) cout << ans[i] << " ";
  cout << endl;
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  for (cin >> T; T--;) solve();
  return 0;
}
