#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2,fma,tune=native")
template <typename T>
void print(T&& x) {
  cout << x << "\n";
}
template <typename T, typename... S>
void print(T&& x, S&&... y) {
  cout << x << ' ';
  print(y...);
}
const long long Mod = 1000000007, Mod2 = 998244353;
const long long MOD = Mod;
const long long maxn = 200005;
long long n, a[maxn];
vector<long long> res;
void op(long long x) {
  long long k = a[x] ^ a[x + 1] ^ a[x + 2];
  a[x] = a[x + 1] = a[x + 2] = k;
  res.push_back(x);
}
void orzck() {
  cin >> n;
  for (long long i = (0); i < ((long long)(n)); i++) cin >> a[i];
  if (accumulate(a, a + n, 0ll) & 1) {
    print("NO");
    return;
  }
  res.clear();
  826;
  long long cnt = 0;
  long long x = -1;
  for (long long i = (0); i < ((long long)(n)); i++) {
    if (a[i])
      cnt++;
    else if (cnt % 2 == 0) {
      x = i;
      break;
    }
  }
  if (x == -1) {
    for (long long i = (0); i < ((long long)(n - 2)); i++) {
      if (a[i + 1] == 0) {
        if (a[i + 2]) {
          op(i);
          x = i;
          break;
        }
        op(i);
      }
    }
  }
  if (x == -1) {
    print("NO");
    return;
  }
  long long l = x;
  while (l < n - 1) {
    if (a[l + 1] == 0) {
      l++;
      continue;
    }
    if (a[l + 2] == 1) {
      op(l);
      l += 2;
      continue;
    }
    if (a[l + 3] == 1) {
      op(l + 1);
      l += 3;
      continue;
    }
    op(l + 1);
    op(l);
    l += 2;
  }
  long long r = x;
  while (r) {
    if (a[r - 1] == 0) {
      r--;
      continue;
    }
    if (a[r - 2] == 1) {
      op(r - 2);
      r -= 2;
      continue;
    }
    if (a[r - 3] == 1) {
      op(r - 3);
      r -= 3;
      continue;
    }
    op(r - 3);
    op(r - 2);
    r -= 2;
  }
  for (long long i = (0); i < ((long long)(n)); i++) 826;
  print("YES");
  print(((long long)res.size()));
  for (auto i : res) cout << (i + 1) << ' ';
  cout << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cout << fixed << setprecision(15);
  long long t;
  cin >> t;
  while (t--) orzck();
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2,fma,tune=native")
template <typename T>
void print(T&& x) {
  cout << x << "\n";
}
template <typename T, typename... S>
void print(T&& x, S&&... y) {
  cout << x << ' ';
  print(y...);
}
const long long Mod = 1000000007, Mod2 = 998244353;
const long long MOD = Mod;
const long long maxn = 200005;
long long n, a[maxn];
vector<long long> res;
void op(long long x) {
  long long k = a[x] ^ a[x + 1] ^ a[x + 2];
  a[x] = a[x + 1] = a[x + 2] = k;
  res.push_back(x);
}
void orzck() {
  cin >> n;
  for (long long i = (0); i < ((long long)(n)); i++) cin >> a[i];
  if (accumulate(a, a + n, 0ll) & 1) {
    print("NO");
    return;
  }
  res.clear();
  826;
  long long cnt = 0;
  long long x = -1;
  for (long long i = (0); i < ((long long)(n)); i++) {
    if (a[i])
      cnt++;
    else if (cnt % 2 == 0) {
      x = i;
      break;
    }
  }
  if (x == -1) {
    for (long long i = (0); i < ((long long)(n - 2)); i++) {
      if (a[i + 1] == 0) {
        if (a[i + 2]) {
          op(i);
          x = i;
          break;
        }
        op(i);
      }
    }
  }
  if (x == -1) {
    print("NO");
    return;
  }
  long long l = x;
  while (l < n - 1) {
    if (a[l + 1] == 0) {
      l++;
      continue;
    }
    if (a[l + 2] == 1) {
      op(l);
      l += 2;
      continue;
    }
    if (a[l + 3] == 1) {
      op(l + 1);
      l += 3;
      continue;
    }
    op(l + 1);
    op(l);
    l += 2;
  }
  long long r = x;
  while (r) {
    if (a[r - 1] == 0) {
      r--;
      continue;
    }
    if (a[r - 2] == 1) {
      op(r - 2);
      r -= 2;
      continue;
    }
    if (a[r - 3] == 1) {
      op(r - 3);
      r -= 3;
      continue;
    }
    op(r - 3);
    op(r - 2);
    r -= 2;
  }
  for (long long i = (0); i < ((long long)(n)); i++) 826;
  print("YES");
  print(((long long)res.size()));
  for (auto i : res) cout << (i + 1) << ' ';
  cout << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cout << fixed << setprecision(15);
  long long t;
  cin >> t;
  while (t--) orzck();
  return 0;
}

```