#include <bits/stdc++.h>
using namespace std;
using vi = vector<long long>;
template <typename T>
std::istream& operator>>(std::istream& input, std::pair<T, T>& data) {
  input >> data.first >> data.second;
  return input;
}
template <typename T>
std::istream& operator>>(std::istream& input, std::vector<T>& data) {
  for (T& first : data) input >> first;
  return input;
}
template <typename T>
std::ostream& operator<<(std::ostream& output, const pair<T, T>& data) {
  output << "(" << data.first << "," << data.second << ")";
  return output;
}
template <typename T>
std::ostream& operator<<(std::ostream& output, const std::vector<T>& data) {
  for (const T& first : data) output << first << " ";
  return output;
}
std::ostream& operator<<(std::ostream& output, const __int128& first) {
  __int128 n = first;
  if (n == 0) {
    output << "0";
    return output;
  }
  if (n < 0) {
    n = -n;
    output << "-";
  }
  string s;
  while (n) {
    s += '0' + (n % 10);
    n /= 10;
  }
  reverse(s.begin(), s.end());
  cout << s;
  return output;
}
long long div_up(long long a, long long b) {
  return a / b + ((a ^ b) > 0 && a % b);
}
long long div_down(long long a, long long b) {
  return a / b - ((a ^ b) < 0 && a % b);
}
long long math_mod(long long a, long long b) { return a - b * div_down(a, b); }
template <class T>
using V = vector<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;
}
long long gcd(long long a, long long b) {
  while (b) {
    tie(a, b) = make_pair(b, a % b);
  }
  return a;
}
long long Bit(long long mask, long long bit) { return (mask >> bit) & 1; }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  for (long long _ = (0); _ < (t); ++_) {
    long long n;
    cin >> n;
    vi a(n);
    cin >> a;
    long long sum = 0;
    for (auto& e : a) {
      sum += e;
    }
    if (sum & 1) {
      cout << "NO" << '\n';
      continue;
    }
    long long pos = -1, l = 0;
    for (long long i = (0); i < (n); ++i) {
      if ((!a[i]) && l % 2 == 0) {
        pos = i;
        break;
      }
      l += a[i];
    }
    auto make = [&](long long mid) {
      a[mid - 1] = a[mid] = a[mid + 1] = a[mid - 1] ^ a[mid] ^ a[mid + 1];
    };
    vi ans;
    if (pos == -1) {
      long long cur = 0;
      for (long long i = (0); i < (n); ++i) {
        if (a[i]) {
          if (cur & 1) {
            for (long long j = 1; j < cur; j += 2) {
              make(i - j);
              ans.push_back(i - j);
            }
            make(i - cur);
            ans.push_back(i - cur);
            pos = i - cur;
            break;
          }
          cur = 0;
        } else {
          cur++;
        }
      }
    };
    if (pos == -1) {
      cout << "NO" << '\n';
      continue;
    }
    auto solve = [&](vi r) {
      vi ans;
      for (long long i = 0; i < r.size(); i += 2) {
        long long p = r[i] + 1;
        while (p + 1 < r[i + 1]) {
          ans.push_back(p);
          p += 2;
        }
        if (p == r[i + 1]) {
          for (long long j = r[i]; j < r[i + 1]; j += 2) {
            ans.push_back(j);
          }
        } else {
          for (long long j = r[i + 1] - 1; j >= r[i]; j -= 2) {
            ans.push_back(j);
          }
        }
      }
      return ans;
    };
    vi r;
    for (long long i = pos + 1; i < n; ++i) {
      if (a[i]) {
        r.push_back(i - pos);
      }
    }
    vi lp;
    for (long long i = pos - 1; i >= 0; --i) {
      if (a[i]) {
        lp.push_back(pos - i);
      }
    }
    for (auto& e : solve(r)) {
      ans.push_back(e + pos);
    }
    for (auto& e : solve(lp)) {
      ans.push_back(pos - e);
    }
    assert(ans.size() <= n);
    cout << "YES" << '\n' << ans.size() << '\n' << ans << '\n';
  }
}

### Prompt

Create a solution in CPP for 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;
using vi = vector<long long>;
template <typename T>
std::istream& operator>>(std::istream& input, std::pair<T, T>& data) {
  input >> data.first >> data.second;
  return input;
}
template <typename T>
std::istream& operator>>(std::istream& input, std::vector<T>& data) {
  for (T& first : data) input >> first;
  return input;
}
template <typename T>
std::ostream& operator<<(std::ostream& output, const pair<T, T>& data) {
  output << "(" << data.first << "," << data.second << ")";
  return output;
}
template <typename T>
std::ostream& operator<<(std::ostream& output, const std::vector<T>& data) {
  for (const T& first : data) output << first << " ";
  return output;
}
std::ostream& operator<<(std::ostream& output, const __int128& first) {
  __int128 n = first;
  if (n == 0) {
    output << "0";
    return output;
  }
  if (n < 0) {
    n = -n;
    output << "-";
  }
  string s;
  while (n) {
    s += '0' + (n % 10);
    n /= 10;
  }
  reverse(s.begin(), s.end());
  cout << s;
  return output;
}
long long div_up(long long a, long long b) {
  return a / b + ((a ^ b) > 0 && a % b);
}
long long div_down(long long a, long long b) {
  return a / b - ((a ^ b) < 0 && a % b);
}
long long math_mod(long long a, long long b) { return a - b * div_down(a, b); }
template <class T>
using V = vector<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;
}
long long gcd(long long a, long long b) {
  while (b) {
    tie(a, b) = make_pair(b, a % b);
  }
  return a;
}
long long Bit(long long mask, long long bit) { return (mask >> bit) & 1; }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  for (long long _ = (0); _ < (t); ++_) {
    long long n;
    cin >> n;
    vi a(n);
    cin >> a;
    long long sum = 0;
    for (auto& e : a) {
      sum += e;
    }
    if (sum & 1) {
      cout << "NO" << '\n';
      continue;
    }
    long long pos = -1, l = 0;
    for (long long i = (0); i < (n); ++i) {
      if ((!a[i]) && l % 2 == 0) {
        pos = i;
        break;
      }
      l += a[i];
    }
    auto make = [&](long long mid) {
      a[mid - 1] = a[mid] = a[mid + 1] = a[mid - 1] ^ a[mid] ^ a[mid + 1];
    };
    vi ans;
    if (pos == -1) {
      long long cur = 0;
      for (long long i = (0); i < (n); ++i) {
        if (a[i]) {
          if (cur & 1) {
            for (long long j = 1; j < cur; j += 2) {
              make(i - j);
              ans.push_back(i - j);
            }
            make(i - cur);
            ans.push_back(i - cur);
            pos = i - cur;
            break;
          }
          cur = 0;
        } else {
          cur++;
        }
      }
    };
    if (pos == -1) {
      cout << "NO" << '\n';
      continue;
    }
    auto solve = [&](vi r) {
      vi ans;
      for (long long i = 0; i < r.size(); i += 2) {
        long long p = r[i] + 1;
        while (p + 1 < r[i + 1]) {
          ans.push_back(p);
          p += 2;
        }
        if (p == r[i + 1]) {
          for (long long j = r[i]; j < r[i + 1]; j += 2) {
            ans.push_back(j);
          }
        } else {
          for (long long j = r[i + 1] - 1; j >= r[i]; j -= 2) {
            ans.push_back(j);
          }
        }
      }
      return ans;
    };
    vi r;
    for (long long i = pos + 1; i < n; ++i) {
      if (a[i]) {
        r.push_back(i - pos);
      }
    }
    vi lp;
    for (long long i = pos - 1; i >= 0; --i) {
      if (a[i]) {
        lp.push_back(pos - i);
      }
    }
    for (auto& e : solve(r)) {
      ans.push_back(e + pos);
    }
    for (auto& e : solve(lp)) {
      ans.push_back(pos - e);
    }
    assert(ans.size() <= n);
    cout << "YES" << '\n' << ans.size() << '\n' << ans << '\n';
  }
}

```