#include <bits/stdc++.h>
#pragma GCC optimize("-Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long MOD = 1e9 + 7;
double eps = 1e-12;
long long mpow(long long a, long long b) {
  if (a == 0) return 0;
  if (b == 0) return 1;
  long long t1 = mpow(a, b / 2);
  t1 *= t1;
  t1 %= MOD;
  if (b % 2) t1 *= a;
  t1 %= MOD;
  return t1;
}
long long mpow(long long a, long long b, long long p) {
  if (a == 0) return 0;
  if (b == 0) return 1;
  long long t1 = mpow(a, b / 2, p);
  t1 *= t1;
  t1 %= p;
  if (b % 2) t1 *= a;
  t1 %= p;
  return t1;
}
long long modinverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
long long range(long long l, long long r) { return l + mt() % (r - l + 1); }
long long rev(long long v) { return mpow(v, MOD - 2); }
long long nc2(long long n) { return (n * (n - 1)) / 2; }
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  long long n0 = 0, n1 = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1)
      n1++;
    else
      n0++;
  }
  if (n0 == 0 || n1 % 2) {
    cout << "NO" << '\n';
    return;
  }
  vector<long long> ans;
  n1 = 0;
  for (long long i = 0; i < n - 2; i++) {
    if (a[i] == 1) n1 ^= 1;
    if (a[i] == 1 && a[i + 1] == 0 && a[i + 2] == 0 && n1 == 1) {
      ans.push_back(i + 1);
      a[i + 1] = a[i + 2] = 1;
      n0 -= 2;
    }
  }
  if (n0 == 0) {
    cout << "NO" << '\n';
    return;
  }
  cout << "YES" << '\n';
  vector<long long> vec;
  for (long long i = 0; i < n; i++) {
    if (a[i] == 1) {
      vec.push_back(i + 1);
    }
    while (((long long)(vec).size()) && (((long long)(vec).size()) % 2 == 0)) {
      if (vec.back() != i + 1) {
        vec.pop_back();
        ans.push_back(vec.back());
        vec.pop_back();
      } else if (vec[((long long)(vec).size()) - 2] != i) {
        vec.pop_back();
        ans.push_back(vec.back());
        vec.pop_back();
      } else if (((long long)(vec).size()) > 2 &&
                 vec[((long long)(vec).size()) - 3] != i - 1) {
        vec.pop_back();
        ans.push_back(vec.back() - 1);
        vec.pop_back();
      } else if (((long long)(vec).size()) == 2 && i > 1) {
        vec.pop_back();
        ans.push_back(vec.back() - 1);
        vec.pop_back();
      } else
        break;
    }
  }
  cout << ((long long)(ans).size()) << '\n';
  for (auto it : ans) cout << it << " ";
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    solve();
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
#pragma GCC optimize("-Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long MOD = 1e9 + 7;
double eps = 1e-12;
long long mpow(long long a, long long b) {
  if (a == 0) return 0;
  if (b == 0) return 1;
  long long t1 = mpow(a, b / 2);
  t1 *= t1;
  t1 %= MOD;
  if (b % 2) t1 *= a;
  t1 %= MOD;
  return t1;
}
long long mpow(long long a, long long b, long long p) {
  if (a == 0) return 0;
  if (b == 0) return 1;
  long long t1 = mpow(a, b / 2, p);
  t1 *= t1;
  t1 %= p;
  if (b % 2) t1 *= a;
  t1 %= p;
  return t1;
}
long long modinverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
long long range(long long l, long long r) { return l + mt() % (r - l + 1); }
long long rev(long long v) { return mpow(v, MOD - 2); }
long long nc2(long long n) { return (n * (n - 1)) / 2; }
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  long long n0 = 0, n1 = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1)
      n1++;
    else
      n0++;
  }
  if (n0 == 0 || n1 % 2) {
    cout << "NO" << '\n';
    return;
  }
  vector<long long> ans;
  n1 = 0;
  for (long long i = 0; i < n - 2; i++) {
    if (a[i] == 1) n1 ^= 1;
    if (a[i] == 1 && a[i + 1] == 0 && a[i + 2] == 0 && n1 == 1) {
      ans.push_back(i + 1);
      a[i + 1] = a[i + 2] = 1;
      n0 -= 2;
    }
  }
  if (n0 == 0) {
    cout << "NO" << '\n';
    return;
  }
  cout << "YES" << '\n';
  vector<long long> vec;
  for (long long i = 0; i < n; i++) {
    if (a[i] == 1) {
      vec.push_back(i + 1);
    }
    while (((long long)(vec).size()) && (((long long)(vec).size()) % 2 == 0)) {
      if (vec.back() != i + 1) {
        vec.pop_back();
        ans.push_back(vec.back());
        vec.pop_back();
      } else if (vec[((long long)(vec).size()) - 2] != i) {
        vec.pop_back();
        ans.push_back(vec.back());
        vec.pop_back();
      } else if (((long long)(vec).size()) > 2 &&
                 vec[((long long)(vec).size()) - 3] != i - 1) {
        vec.pop_back();
        ans.push_back(vec.back() - 1);
        vec.pop_back();
      } else if (((long long)(vec).size()) == 2 && i > 1) {
        vec.pop_back();
        ans.push_back(vec.back() - 1);
        vec.pop_back();
      } else
        break;
    }
  }
  cout << ((long long)(ans).size()) << '\n';
  for (auto it : ans) cout << it << " ";
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    solve();
  }
  return 0;
}

```