#include <bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
clock_t startTime;
double getCurrentTime() {
  return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  int totalXor = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    totalXor ^= a[i];
  }
  if (totalXor == 1) {
    cout << "NO\n";
    return;
  }
  auto solve = [](int l, int r) {
    assert((r - l + 1) % 2 == 1);
    vector<int> res;
    for (int i = l; i <= r - 2; i += 2) {
      res.push_back(i);
    }
    for (int i = r - 4; i >= l; i -= 2) {
      res.push_back(i);
    }
    return res;
  };
  if (n % 2 == 1) {
    auto res = solve(0, n - 1);
    cout << "YES\n";
    cout << res.size() << "\n";
    for (auto x : res) {
      cout << x + 1 << " ";
    }
    cout << "\n";
  } else {
    bool ok = false;
    int prefXor = 0;
    for (int i = 0; i < n; ++i) {
      prefXor ^= a[i];
      if (prefXor == 0 && i % 2 == 0) {
        auto res1 = solve(0, i);
        auto res2 = solve(i + 1, n - 1);
        cout << "YES\n";
        cout << res1.size() + res2.size() << "\n";
        for (auto x : res1) {
          cout << x + 1 << " ";
        }
        for (auto x : res2) {
          cout << x + 1 << " ";
        }
        cout << "\n";
        ok = true;
        break;
      }
    }
    if (!ok) cout << "NO\n";
  }
}
int main() {
  startTime = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int tests;
  cin >> tests;
  for (int t = 0; t < tests; ++t) {
    solve();
  }
  return 0;
}

### 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;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
clock_t startTime;
double getCurrentTime() {
  return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  int totalXor = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    totalXor ^= a[i];
  }
  if (totalXor == 1) {
    cout << "NO\n";
    return;
  }
  auto solve = [](int l, int r) {
    assert((r - l + 1) % 2 == 1);
    vector<int> res;
    for (int i = l; i <= r - 2; i += 2) {
      res.push_back(i);
    }
    for (int i = r - 4; i >= l; i -= 2) {
      res.push_back(i);
    }
    return res;
  };
  if (n % 2 == 1) {
    auto res = solve(0, n - 1);
    cout << "YES\n";
    cout << res.size() << "\n";
    for (auto x : res) {
      cout << x + 1 << " ";
    }
    cout << "\n";
  } else {
    bool ok = false;
    int prefXor = 0;
    for (int i = 0; i < n; ++i) {
      prefXor ^= a[i];
      if (prefXor == 0 && i % 2 == 0) {
        auto res1 = solve(0, i);
        auto res2 = solve(i + 1, n - 1);
        cout << "YES\n";
        cout << res1.size() + res2.size() << "\n";
        for (auto x : res1) {
          cout << x + 1 << " ";
        }
        for (auto x : res2) {
          cout << x + 1 << " ";
        }
        cout << "\n";
        ok = true;
        break;
      }
    }
    if (!ok) cout << "NO\n";
  }
}
int main() {
  startTime = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int tests;
  cin >> tests;
  for (int t = 0; t < tests; ++t) {
    solve();
  }
  return 0;
}

```