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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using Vi = vector<int>;
using i64 = long long;
using ll = long long;
using Pi = pair<int, int>;
mt19937 mrand(time(0));
int rnd(int x) { return mrand() % x; }
const int N = 500005;
const int M = 40005;
int n;
Vi a, ans;
void soll(int x) {
  if (x > 1) {
    if (a[x] + a[x - 1] + a[x - 2] == 2) {
      ans.push_back(x - 2);
      a[x - 1] = a[x - 2] = a[x] = 0;
    }
    soll(x - 1);
  }
};
void solr(int x) {
  if (x < n - 2) {
    if (a[x] + a[x + 1] + a[x + 2] == 2) {
      ans.push_back(x);
      a[x] = a[x + 1] = a[x + 2] = 0;
    }
    solr(x + 1);
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    a.resize(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    int l = 1000000001, r = -1;
    for (int i = 0; i < n; ++i)
      if (a[i]) l = min(l, i), r = max(r, i);
    if (r == -1) {
      cout << "YES" << endl << "0" << endl;
      continue;
    }
    ans.clear();
    for (int i = l; i < r - 1; ++i)
      if (a[i] == 1 && a[i + 1] == 0 && a[i + 2] == 0)
        ans.push_back(i), a[i + 1] = a[i + 2] = 1;
    int zeros = 0;
    for (int i = l; i < r + 1; ++i)
      if (a[i] == 0) ++zeros;
    if ((zeros == 0 && l == 0 && r == n - 1) || ((r - l + 1 - zeros) & 1)) {
      cout << "NO" << endl;
      continue;
    }
    if (a[0] == 0)
      solr(0);
    else if (a[n - 1] == 0)
      soll(n - 1);
    else {
      int ones = 0;
      for (int i = 0; i < n; ++i) {
        if (a[i] == 0) {
          if (ones & 1) {
            ans.push_back(i - 1);
            a[i - 1] = a[i + 1] = 0;
            soll(i - 1);
            solr(i + 1);
          } else {
            ans.push_back(i - 2);
            a[i - 2] = a[i - 1] = 0;
            soll(i - 2);
            solr(i);
          }
          break;
        }
        ++ones;
      }
    }
    cout << "YES" << endl << (int)ans.size() << endl;
    assert((int)ans.size() <= n);
    for (auto i : ans) cout << i + 1 << " ";
    cout << endl;
  }
  return 0;
}
