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 long long INF = 1e18;
long long ceel(long long a, long long b) { return ((a + b - 1) / b); }
bool isprime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long power(long long x, long long y, long long mod) {
  long long ans = 1;
  x = x % mod;
  while (y) {
    if (y % 2 == 1) ans = (ans * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return ans;
}
void solve() {
  long long n, x;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long fx = 0;
  for (long long i = 0; i < n; i++) {
    fx = fx ^ a[i];
  }
  if (fx == 1) {
    cout << "NmO" << '\n';
    return;
  }
  fx = 0;
  if (n % 2) {
    cout << "YES" << '\n';
    vector<long long> v;
    for (long long i = 0; i < n - 2; i += 2) {
      v.push_back(i + 1);
    }
    for (long long i = n - 4; i > 0; i -= 2) {
      v.push_back(i);
    }
    cout << v.size() << '\n';
    for (auto ele : v) {
      cout << ele << " ";
    }
    cout << '\n';
  } else {
    long long j = -1;
    for (long long i = 0; i < n; i++) {
      fx = fx ^ a[i];
      if (fx == 0 && i % 2 == 0) {
        j = i;
        break;
      }
    }
    if (j == -1) {
      cout << "NO" << '\n';
      return;
    }
    vector<long long> v;
    cout << "YES" << '\n';
    for (long long i = 0; i < j; i += 2) {
      v.push_back(i + 1);
    }
    for (long long i = j - 1; i > 0; i -= 2) {
      v.push_back(i);
    }
    for (long long i = j + 1; i < n; i += 2) {
      v.push_back(i + 1);
    }
    for (long long i = n - 2; i > j; i -= 2) {
      v.push_back(i);
    }
    cout << v.size() << '\n';
    for (auto ele : v) {
      cout << ele << " ";
    }
    cout << '\n';
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t;
  cin >> t;
  for (long long tc = 0; tc < t; tc++) solve();
  return 0;
}
