#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <class T>
using pq = priority_queue<T>;
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char* x) { cerr << '\"' << x << '\"'; }
void __print(const string& x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V>& x) {
  cerr << '{';
  __print(x.first);
  cerr << ", ";
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T& x) {
  int first = 0;
  cerr << '{';
  for (auto& i : x) cerr << (first++ ? ", " : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const int MOD = 1000000007;
const char nl = '\n';
const int MX = 100001;
void solve() {
  int N;
  cin >> N;
  int A[N];
  for (int i = 0; i < (N); i++) cin >> A[i];
  int sum = 0;
  for (int i = 0; i < (N); i++) sum += A[i];
  if (sum % 2) {
    cout << "NO" << nl;
    return;
  }
  int cnt = 0;
  vector<int> ops;
  for (int i = 0; i < (N); i++) {
    cnt += A[i];
    if (A[i] == 0) {
      if (cnt % 2) {
        ops.push_back(i - 1);
        int X = 0;
        for (int j = 0; j < (3); j++) X ^= A[i - 1 + j];
        for (int j = 0; j < (3); j++) A[i - 1 + j] = X;
      }
      if (A[i] == 0) {
        cnt = 0;
      } else
        cnt++;
    }
  }
  sum = 0;
  for (int i = 0; i < (N); i++) sum += A[i];
  if (sum == N) {
    cout << "NO" << nl;
    return;
  }
  int fz = 0;
  for (int i = (N)-1; i >= 0; i--)
    if (A[i] == 0) fz = i;
  while (fz > 0) {
    ops.push_back(fz - 2);
    A[fz - 2] = 0;
    A[fz - 1] = 0;
    A[fz] = 0;
    fz -= 2;
  }
  for (int i = 0; i < (N - 1); i++) {
    if (A[i] == 0 && A[i + 1] == 1) {
      ops.push_back(i);
      A[i] = 0;
      A[i + 1] = 0;
      A[i + 2] = 0;
    }
  }
  cout << "YES" << nl;
  cout << (int)(ops).size() << nl;
  for (auto& a : ops) {
    cout << a + 1 << " ";
  }
  cout << nl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

### 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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <class T>
using pq = priority_queue<T>;
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char* x) { cerr << '\"' << x << '\"'; }
void __print(const string& x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V>& x) {
  cerr << '{';
  __print(x.first);
  cerr << ", ";
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T& x) {
  int first = 0;
  cerr << '{';
  for (auto& i : x) cerr << (first++ ? ", " : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const int MOD = 1000000007;
const char nl = '\n';
const int MX = 100001;
void solve() {
  int N;
  cin >> N;
  int A[N];
  for (int i = 0; i < (N); i++) cin >> A[i];
  int sum = 0;
  for (int i = 0; i < (N); i++) sum += A[i];
  if (sum % 2) {
    cout << "NO" << nl;
    return;
  }
  int cnt = 0;
  vector<int> ops;
  for (int i = 0; i < (N); i++) {
    cnt += A[i];
    if (A[i] == 0) {
      if (cnt % 2) {
        ops.push_back(i - 1);
        int X = 0;
        for (int j = 0; j < (3); j++) X ^= A[i - 1 + j];
        for (int j = 0; j < (3); j++) A[i - 1 + j] = X;
      }
      if (A[i] == 0) {
        cnt = 0;
      } else
        cnt++;
    }
  }
  sum = 0;
  for (int i = 0; i < (N); i++) sum += A[i];
  if (sum == N) {
    cout << "NO" << nl;
    return;
  }
  int fz = 0;
  for (int i = (N)-1; i >= 0; i--)
    if (A[i] == 0) fz = i;
  while (fz > 0) {
    ops.push_back(fz - 2);
    A[fz - 2] = 0;
    A[fz - 1] = 0;
    A[fz] = 0;
    fz -= 2;
  }
  for (int i = 0; i < (N - 1); i++) {
    if (A[i] == 0 && A[i + 1] == 1) {
      ops.push_back(i);
      A[i] = 0;
      A[i + 1] = 0;
      A[i + 2] = 0;
    }
  }
  cout << "YES" << nl;
  cout << (int)(ops).size() << nl;
  for (auto& a : ops) {
    cout << a + 1 << " ";
  }
  cout << nl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

```