#include <bits/stdc++.h>
using namespace std;
mt19937 rng((long long)chrono::steady_clock::now().time_since_epoch().count());
const long long mod = 1e9 + 7;
long long n, a[200010];
void op_ida(long long Ini, vector<long long> &ans) {
  for (long long ini = Ini; ini <= n - 2; ini++) {
    long long meio = ini + 1, fim = ini + 2;
    if (a[meio] == 0) continue;
    if (a[fim] == 1) {
      ans.push_back(ini);
      a[meio] = a[fim] = 0;
      continue;
    }
    if (meio <= n - 2) {
      if (a[fim + 1] == 1) {
        ans.push_back(meio);
        a[meio] = a[fim + 1] = 0;
      } else {
        ans.push_back(meio);
        ans.push_back(ini);
        a[meio] = a[fim] = 0;
        a[fim + 1] = 1;
      }
      continue;
    }
    cout << "NO\n";
    exit(0);
  }
}
void op_volta(long long Ini, vector<long long> &ans) {
  for (long long ini = Ini; ini >= 3; ini--) {
    long long meio = ini - 1, fim = ini - 2;
    if (a[meio] == 0) continue;
    if (a[fim] == 1) {
      ans.push_back(fim);
      a[meio] = a[fim] = 0;
      continue;
    }
    if (meio >= 3) {
      if (a[fim - 1] == 1) {
        ans.push_back(fim - 1);
        a[meio] = a[fim - 1] = 0;
      } else {
        ans.push_back(fim - 1);
        ans.push_back(fim);
        a[meio] = a[fim] = 0;
        a[fim - 1] = 1;
      }
      continue;
    }
    cout << "NO\n";
    exit(0);
  }
}
void solve() {
  cin >> n;
  a[n + 1] = a[0] = 0;
  long long tot = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    tot += a[i];
  }
  if (tot & 1 or tot == n) {
    cout << "NO\n";
    return;
  }
  if (a[1] == 0) {
    vector<long long> ans;
    op_ida(1, ans);
    cout << "YES\n";
    cout << ans.size() << "\n";
    for (auto u : ans) cout << u << " ";
    cout << "\n";
    return;
  }
  if (a[n] == 0) {
    vector<long long> ans;
    op_volta(n, ans);
    cout << "YES\n";
    cout << ans.size() << "\n";
    for (auto u : ans) cout << u << " ";
    cout << "\n";
    return;
  }
  long long pref = 0;
  for (long long i = 1; i <= n; i++) {
    pref += a[i];
    if (a[i] == 1) continue;
    if (pref % 2 == 0) {
      vector<long long> ans;
      op_ida(i, ans);
      op_volta(i, ans);
      cout << "YES\n";
      cout << ans.size() << "\n";
      for (auto u : ans) cout << u << " ";
      cout << "\n";
      return;
    } else {
      long long j = i;
      while (j <= n and a[j] == 0) j++;
      j--;
      if ((j - i + 1) & 1) {
        vector<long long> ans;
        while (j != i) {
          a[j] = a[j - 1] = 1;
          ans.push_back(j - 1);
          j -= 2;
        }
        ans.push_back(i - 1);
        a[i + 1] = 0;
        a[i - 1] = 0;
        op_ida(i, ans);
        op_volta(i, ans);
        cout << "YES\n";
        cout << ans.size() << "\n";
        for (auto u : ans) cout << u << " ";
        cout << "\n";
        return;
      }
      i = j;
    }
  }
  cout << "NO\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  for (long long i = 1; i <= t; i++) solve();
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 rng((long long)chrono::steady_clock::now().time_since_epoch().count());
const long long mod = 1e9 + 7;
long long n, a[200010];
void op_ida(long long Ini, vector<long long> &ans) {
  for (long long ini = Ini; ini <= n - 2; ini++) {
    long long meio = ini + 1, fim = ini + 2;
    if (a[meio] == 0) continue;
    if (a[fim] == 1) {
      ans.push_back(ini);
      a[meio] = a[fim] = 0;
      continue;
    }
    if (meio <= n - 2) {
      if (a[fim + 1] == 1) {
        ans.push_back(meio);
        a[meio] = a[fim + 1] = 0;
      } else {
        ans.push_back(meio);
        ans.push_back(ini);
        a[meio] = a[fim] = 0;
        a[fim + 1] = 1;
      }
      continue;
    }
    cout << "NO\n";
    exit(0);
  }
}
void op_volta(long long Ini, vector<long long> &ans) {
  for (long long ini = Ini; ini >= 3; ini--) {
    long long meio = ini - 1, fim = ini - 2;
    if (a[meio] == 0) continue;
    if (a[fim] == 1) {
      ans.push_back(fim);
      a[meio] = a[fim] = 0;
      continue;
    }
    if (meio >= 3) {
      if (a[fim - 1] == 1) {
        ans.push_back(fim - 1);
        a[meio] = a[fim - 1] = 0;
      } else {
        ans.push_back(fim - 1);
        ans.push_back(fim);
        a[meio] = a[fim] = 0;
        a[fim - 1] = 1;
      }
      continue;
    }
    cout << "NO\n";
    exit(0);
  }
}
void solve() {
  cin >> n;
  a[n + 1] = a[0] = 0;
  long long tot = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    tot += a[i];
  }
  if (tot & 1 or tot == n) {
    cout << "NO\n";
    return;
  }
  if (a[1] == 0) {
    vector<long long> ans;
    op_ida(1, ans);
    cout << "YES\n";
    cout << ans.size() << "\n";
    for (auto u : ans) cout << u << " ";
    cout << "\n";
    return;
  }
  if (a[n] == 0) {
    vector<long long> ans;
    op_volta(n, ans);
    cout << "YES\n";
    cout << ans.size() << "\n";
    for (auto u : ans) cout << u << " ";
    cout << "\n";
    return;
  }
  long long pref = 0;
  for (long long i = 1; i <= n; i++) {
    pref += a[i];
    if (a[i] == 1) continue;
    if (pref % 2 == 0) {
      vector<long long> ans;
      op_ida(i, ans);
      op_volta(i, ans);
      cout << "YES\n";
      cout << ans.size() << "\n";
      for (auto u : ans) cout << u << " ";
      cout << "\n";
      return;
    } else {
      long long j = i;
      while (j <= n and a[j] == 0) j++;
      j--;
      if ((j - i + 1) & 1) {
        vector<long long> ans;
        while (j != i) {
          a[j] = a[j - 1] = 1;
          ans.push_back(j - 1);
          j -= 2;
        }
        ans.push_back(i - 1);
        a[i + 1] = 0;
        a[i - 1] = 0;
        op_ida(i, ans);
        op_volta(i, ans);
        cout << "YES\n";
        cout << ans.size() << "\n";
        for (auto u : ans) cout << u << " ";
        cout << "\n";
        return;
      }
      i = j;
    }
  }
  cout << "NO\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  for (long long i = 1; i <= t; i++) solve();
}

```