#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long po(long long x) { return ((x % mod) + mod) % mod; }
long long SM(long long x, long long y) { return (po(x) + po(y)) % mod; }
long long M(long long x, long long y) { return (po(x) * po(y)) % mod; }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long x, long long y) {
  return uniform_int_distribution<long long>(x, y)(rng);
}
long long t = 1, n, i, j, k, x, y, z, l, r, mid, m, ans;
long long a[400005], b[400005], c[400005], d[400005];
map<long long, long long> mp;
string s, st;
vector<long long> v[300005], u;
long long ff() {
  x = 0;
  u.clear();
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i], x += a[i];
  if (x % 2 || x == n) return cout << "NO", 0;
  if (x == 0) return cout << "YES\n0", 0;
  if (a[0]) {
    for (i = 0; i + 2 < n; i++)
      if (a[i] + a[i + 1] + a[i + 2] == 1) {
        a[i] = a[i + 1] = a[i + 2] = 1;
        u.push_back(i + 1);
      } else if (a[i] + a[i + 1] + a[i + 2] == 2 && i % 2 == 0) {
        a[i] = a[i + 1] = a[i + 2] = 0;
        u.push_back(i + 1);
        break;
      }
    if (i + 2 == n) return cout << "NO", 0;
    while (i >= 0) {
      if (a[i] + a[i + 1] + a[i + 2] == 2) {
        a[i] = a[i + 1] = a[i + 2] = 0;
        u.push_back(i + 1);
      }
      i--;
    }
  }
  for (i = 0; i + 2 < n; i++)
    if (a[i] + a[i + 1] + a[i + 2] == 2) {
      u.push_back(i + 1);
      a[i] = a[i + 1] = a[i + 2] = 0;
    } else if (a[i] && a[i + 1] + a[i + 2] == 0) {
      u.push_back(i + 1);
      u.push_back(i);
      swap(a[i], a[i + 2]);
    }
  cout << "YES";
  cout << "\n" << u.size() << "\n";
  for (auto i : u) cout << i << " ";
  return 0;
}
int main() {
  ios::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    ff();
    cout << "\n";
  }
}

### Prompt

In cpp, your task is to solve 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;
long long mod = 1e9 + 7;
long long po(long long x) { return ((x % mod) + mod) % mod; }
long long SM(long long x, long long y) { return (po(x) + po(y)) % mod; }
long long M(long long x, long long y) { return (po(x) * po(y)) % mod; }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long x, long long y) {
  return uniform_int_distribution<long long>(x, y)(rng);
}
long long t = 1, n, i, j, k, x, y, z, l, r, mid, m, ans;
long long a[400005], b[400005], c[400005], d[400005];
map<long long, long long> mp;
string s, st;
vector<long long> v[300005], u;
long long ff() {
  x = 0;
  u.clear();
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i], x += a[i];
  if (x % 2 || x == n) return cout << "NO", 0;
  if (x == 0) return cout << "YES\n0", 0;
  if (a[0]) {
    for (i = 0; i + 2 < n; i++)
      if (a[i] + a[i + 1] + a[i + 2] == 1) {
        a[i] = a[i + 1] = a[i + 2] = 1;
        u.push_back(i + 1);
      } else if (a[i] + a[i + 1] + a[i + 2] == 2 && i % 2 == 0) {
        a[i] = a[i + 1] = a[i + 2] = 0;
        u.push_back(i + 1);
        break;
      }
    if (i + 2 == n) return cout << "NO", 0;
    while (i >= 0) {
      if (a[i] + a[i + 1] + a[i + 2] == 2) {
        a[i] = a[i + 1] = a[i + 2] = 0;
        u.push_back(i + 1);
      }
      i--;
    }
  }
  for (i = 0; i + 2 < n; i++)
    if (a[i] + a[i + 1] + a[i + 2] == 2) {
      u.push_back(i + 1);
      a[i] = a[i + 1] = a[i + 2] = 0;
    } else if (a[i] && a[i + 1] + a[i + 2] == 0) {
      u.push_back(i + 1);
      u.push_back(i);
      swap(a[i], a[i + 2]);
    }
  cout << "YES";
  cout << "\n" << u.size() << "\n";
  for (auto i : u) cout << i << " ";
  return 0;
}
int main() {
  ios::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    ff();
    cout << "\n";
  }
}

```