#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
unsigned seed = chrono::system_clock::now().time_since_epoch().count();
mt19937 eng(seed);
long long random2() { return (1ll << 31ll) * eng() + eng(); }
long long n, m, k, q, T;
const long long big = 1000000007;
const long long big2 = 1000000009;
const long long mod = 998244353;
const long long MAXN = 400004;
vector<int> A;
vector<int> ANS;
void mov(long long i) {
  ANS.push_back(i);
  long long x = (A[i] ^ A[i + 1] ^ A[i + 2]);
  for (int c1 = 0; c1 < (3); ++c1) {
    A[i + c1] = x;
  }
}
void even(long long i, long long j) {
  if (j == n - 1) {
    for (int c1 = i - 1; c1 < j; c1 += 2) {
      mov(c1);
    }
  } else {
    for (int c1 = j - 1; c1 >= i; c1 -= 2) {
      mov(c1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, c, d, e;
  cin >> T;
  for (int c4 = 0; c4 < (T); ++c4) {
    cin >> n;
    A.clear();
    ANS.clear();
    b = 0;
    for (int c1 = 0; c1 < (n); ++c1) {
      cin >> a;
      A.push_back(a);
      b += a;
    }
    if (b % 2 == 1 || b == n) {
      cout << "NO\n";
      continue;
    }
    long long start = -1;
    for (int c1 = 0; c1 < (n); ++c1) {
      if (A[c1] == 0) {
        if (start == -1) continue;
        int parity = (c1 - start) % 2;
        if (parity == 0) {
          even(start, c1 - 1);
          start = -1;
        } else {
          if (A[c1 + 1] == 1) {
            mov(c1 - 1);
            even(start, c1 - 2);
            start = -1;
          } else {
            mov(c1 - 1);
          }
        }
      } else {
        if (start == -1) {
          start = c1;
        }
      }
    }
    start = -1;
    for (int c1 = 0; c1 < (n); ++c1) {
      if (A[c1] == 1) {
        start = c1;
        break;
      }
    }
    if (start != -1) {
      if (start == 0) {
        cout << "NO\n";
        continue;
      } else {
        even(start, n - 1);
      }
    }
    cout << "YES\n";
    cout << (int)(ANS).size() << "\n";
    for (auto& y : ANS) {
      cout << y + 1 << " ";
    }
    cout << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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("O3")
using namespace std;
unsigned seed = chrono::system_clock::now().time_since_epoch().count();
mt19937 eng(seed);
long long random2() { return (1ll << 31ll) * eng() + eng(); }
long long n, m, k, q, T;
const long long big = 1000000007;
const long long big2 = 1000000009;
const long long mod = 998244353;
const long long MAXN = 400004;
vector<int> A;
vector<int> ANS;
void mov(long long i) {
  ANS.push_back(i);
  long long x = (A[i] ^ A[i + 1] ^ A[i + 2]);
  for (int c1 = 0; c1 < (3); ++c1) {
    A[i + c1] = x;
  }
}
void even(long long i, long long j) {
  if (j == n - 1) {
    for (int c1 = i - 1; c1 < j; c1 += 2) {
      mov(c1);
    }
  } else {
    for (int c1 = j - 1; c1 >= i; c1 -= 2) {
      mov(c1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, c, d, e;
  cin >> T;
  for (int c4 = 0; c4 < (T); ++c4) {
    cin >> n;
    A.clear();
    ANS.clear();
    b = 0;
    for (int c1 = 0; c1 < (n); ++c1) {
      cin >> a;
      A.push_back(a);
      b += a;
    }
    if (b % 2 == 1 || b == n) {
      cout << "NO\n";
      continue;
    }
    long long start = -1;
    for (int c1 = 0; c1 < (n); ++c1) {
      if (A[c1] == 0) {
        if (start == -1) continue;
        int parity = (c1 - start) % 2;
        if (parity == 0) {
          even(start, c1 - 1);
          start = -1;
        } else {
          if (A[c1 + 1] == 1) {
            mov(c1 - 1);
            even(start, c1 - 2);
            start = -1;
          } else {
            mov(c1 - 1);
          }
        }
      } else {
        if (start == -1) {
          start = c1;
        }
      }
    }
    start = -1;
    for (int c1 = 0; c1 < (n); ++c1) {
      if (A[c1] == 1) {
        start = c1;
        break;
      }
    }
    if (start != -1) {
      if (start == 0) {
        cout << "NO\n";
        continue;
      } else {
        even(start, n - 1);
      }
    }
    cout << "YES\n";
    cout << (int)(ANS).size() << "\n";
    for (auto& y : ANS) {
      cout << y + 1 << " ";
    }
    cout << "\n";
  }
  return 0;
}

```