#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 200005;
int n;
int a[N];
int p[N], s[N];
vector<int> ans;
void ubd(int i) {
  assert(1 <= i && i <= n - 2);
  ans.push_back(i);
  int x = (a[i] ^ a[i + 1] ^ a[i + 2]);
  a[i] = a[i + 1] = a[i + 2] = x;
}
bool solv() {
  cin >> n;
  for (int i = 0; i <= n + 1; ++i) {
    p[i] = s[i] = 0;
  }
  ans.clear();
  for (int i = 1; i <= n; ++i) cin >> a[i];
  bool z = false;
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 0) {
      z = true;
      break;
    }
  }
  if (!z) {
    cout << "NO\n";
    return false;
  }
  for (int i = 1; i <= n; ++i) p[i] = p[i - 1] + a[i];
  for (int i = n; i >= 1; --i) s[i] = s[i + 1] + a[i];
  if (p[n] % 2 == 1) {
    cout << "NO\n";
    return false;
  }
  int u = -1;
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 0 && p[i] % 2 == 0 && s[i] % 2 == 0) {
      u = i;
      break;
    }
  }
  if (u == -1) {
    int l, r;
    bool f = false;
    for (int i = 1; i <= n; ++i) {
      if (a[i] == 0) {
        if (i == 0 || a[i - 1] == 1) {
          l = i;
        }
        r = i;
        if (i == n || a[i + 1] == 1) {
          if ((r - l + 1) % 2 == 1) {
            f = true;
            break;
          }
        }
      }
    }
    if (!f) {
      cout << "NO\n";
      return false;
    }
    for (int i = l - 1; i <= n; ++i) {
      if (a[i] == 1 && a[i + 1] == 0 && a[i + 2] == 1) {
        ubd(i);
        u = i;
        break;
      } else if (a[i] == 1 && a[i + 1] == 0 && a[i + 2] == 0) {
        ubd(i);
      }
    }
  }
  if (u != -1) {
    for (int i = u; i <= n; ++i) {
      if (a[i] == 1) {
        if (a[i + 1] == 1) {
          ubd(i - 1);
        } else if (a[i + 2] == 1) {
          ubd(i);
        } else {
          ubd(i);
          ubd(i - 1);
        }
      }
    }
    for (int i = u; i >= 1; --i) {
      if (a[i] == 1) {
        if (a[i - 1] == 1) {
          ubd(i - 1);
        } else if (a[i - 2] == 1) {
          ubd(i - 2);
        } else {
          ubd(i - 2);
          ubd(i - 1);
        }
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    assert(a[i] == 0);
  }
  assert(((int)(ans).size()) <= n);
  cout << "YES\n";
  cout << ((int)(ans).size()) << "\n";
  if (!ans.empty()) {
    for (int i = 0; i < ((int)(ans).size()); ++i) cout << ans[i] << ' ';
    cout << "\n";
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int tt = 1;
  cin >> tt;
  while (tt--) solv();
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 200005;
int n;
int a[N];
int p[N], s[N];
vector<int> ans;
void ubd(int i) {
  assert(1 <= i && i <= n - 2);
  ans.push_back(i);
  int x = (a[i] ^ a[i + 1] ^ a[i + 2]);
  a[i] = a[i + 1] = a[i + 2] = x;
}
bool solv() {
  cin >> n;
  for (int i = 0; i <= n + 1; ++i) {
    p[i] = s[i] = 0;
  }
  ans.clear();
  for (int i = 1; i <= n; ++i) cin >> a[i];
  bool z = false;
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 0) {
      z = true;
      break;
    }
  }
  if (!z) {
    cout << "NO\n";
    return false;
  }
  for (int i = 1; i <= n; ++i) p[i] = p[i - 1] + a[i];
  for (int i = n; i >= 1; --i) s[i] = s[i + 1] + a[i];
  if (p[n] % 2 == 1) {
    cout << "NO\n";
    return false;
  }
  int u = -1;
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 0 && p[i] % 2 == 0 && s[i] % 2 == 0) {
      u = i;
      break;
    }
  }
  if (u == -1) {
    int l, r;
    bool f = false;
    for (int i = 1; i <= n; ++i) {
      if (a[i] == 0) {
        if (i == 0 || a[i - 1] == 1) {
          l = i;
        }
        r = i;
        if (i == n || a[i + 1] == 1) {
          if ((r - l + 1) % 2 == 1) {
            f = true;
            break;
          }
        }
      }
    }
    if (!f) {
      cout << "NO\n";
      return false;
    }
    for (int i = l - 1; i <= n; ++i) {
      if (a[i] == 1 && a[i + 1] == 0 && a[i + 2] == 1) {
        ubd(i);
        u = i;
        break;
      } else if (a[i] == 1 && a[i + 1] == 0 && a[i + 2] == 0) {
        ubd(i);
      }
    }
  }
  if (u != -1) {
    for (int i = u; i <= n; ++i) {
      if (a[i] == 1) {
        if (a[i + 1] == 1) {
          ubd(i - 1);
        } else if (a[i + 2] == 1) {
          ubd(i);
        } else {
          ubd(i);
          ubd(i - 1);
        }
      }
    }
    for (int i = u; i >= 1; --i) {
      if (a[i] == 1) {
        if (a[i - 1] == 1) {
          ubd(i - 1);
        } else if (a[i - 2] == 1) {
          ubd(i - 2);
        } else {
          ubd(i - 2);
          ubd(i - 1);
        }
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    assert(a[i] == 0);
  }
  assert(((int)(ans).size()) <= n);
  cout << "YES\n";
  cout << ((int)(ans).size()) << "\n";
  if (!ans.empty()) {
    for (int i = 0; i < ((int)(ans).size()); ++i) cout << ans[i] << ' ';
    cout << "\n";
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int tt = 1;
  cin >> tt;
  while (tt--) solv();
  return 0;
}

```