#include <bits/stdc++.h>
using namespace std;
int a[200001];
void op(int i) {
  int res = a[i] ^ a[i + 1] ^ a[i + 2];
  a[i] = a[i + 1] = a[i + 2] = res;
}
vector<int> push_left(int i) {
  vector<int> v;
  while (i - 2 >= 1 && a[i] == 0 && a[i - 1] == 1 && a[i - 2] == 1) {
    op(i - 2);
    v.push_back(i - 2);
    i -= 2;
  }
  return v;
}
vector<int> push_right(int i, int n) {
  vector<int> v;
  while (i + 2 <= n && a[i] == 0 && a[i + 1] == 1 && a[i + 2] == 1) {
    op(i);
    v.push_back(i);
    i += 2;
  }
  return v;
}
int main() {
  int i, t;
  for (scanf("%d", &t); t--;) {
    int n;
    vector<int> v, ans;
    int cum = 0;
    for (scanf("%d", &n), i = 1; i <= n; ++i) {
      scanf("%d", a + i);
      cum += a[i];
    }
    if (cum == 0) {
      printf("YES\n0\n");
      continue;
    } else if ((cum & 1) || cum == n) {
      printf("NO\n");
      continue;
    }
    int cur = 1;
    if (a[1]) {
      cum = a[1];
      cur = 2;
      while (cur <= n && a[cur]) cum += a[cur], ++cur;
      if (cum & 1) {
        --cur;
        bool y = 1;
        while (cur <= n && a[cur]) {
          if (cur + 2 <= n) {
            if (a[cur] + a[cur + 1] + a[cur + 2] == 3) {
              cur += 2;
            } else if (a[cur + 1] + a[cur + 2] == 0) {
              ans.push_back(cur);
              op(cur);
              cur += 2;
            } else if (a[cur + 1] == 0 && a[cur + 2] == 1) {
              ans.push_back(cur);
              op(cur);
              auto push_vec = push_left(cur);
              ans.insert(ans.end(), push_vec.begin(), push_vec.end());
              cur += 3;
              break;
            } else {
              auto push_vec = push_left(cur + 2);
              ans.insert(ans.end(), push_vec.begin(), push_vec.end());
              cur += 3;
              break;
            }
          } else {
            y = 0;
            break;
          }
        }
        if (y == 0) {
          printf("NO\n");
          continue;
        }
      } else {
        ans = push_left(cur);
      }
    }
    cur = 1;
    while (cur + 2 <= n) {
      if (a[cur] == 0 && a[cur + 1] + a[cur + 2] == 2) {
        op(cur);
        ans.push_back(cur);
        cur += 2;
      } else if (a[cur]) {
        int st = cur - 1;
        while (1) {
          if (cur + 2 <= n && a[cur + 1] + a[cur + 2] == 2)
            cur += 2;
          else if (cur + 2 <= n && a[cur + 1] + a[cur + 2] == 0) {
            op(cur);
            ans.push_back(cur);
            cur += 2;
          } else if (cur + 2 <= n && a[cur + 1] == 0 && a[cur + 2] == 1) {
            op(cur);
            ans.push_back(cur);
            break;
          } else {
            break;
          }
        }
        auto push_vec = push_right(st, n);
        ans.insert(ans.end(), push_vec.begin(), push_vec.end());
      } else
        ++cur;
    }
    cum = 0;
    for (i = 1; i <= n; ++i) cum += a[i];
    if (cum == 0) {
      printf("YES\n%d\n", ans.size());
      for (auto x : ans) printf("%d ", x);
      puts("");
    } else
      printf("NO\n");
  }
  return 0;
}

### Prompt

Generate 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;
int a[200001];
void op(int i) {
  int res = a[i] ^ a[i + 1] ^ a[i + 2];
  a[i] = a[i + 1] = a[i + 2] = res;
}
vector<int> push_left(int i) {
  vector<int> v;
  while (i - 2 >= 1 && a[i] == 0 && a[i - 1] == 1 && a[i - 2] == 1) {
    op(i - 2);
    v.push_back(i - 2);
    i -= 2;
  }
  return v;
}
vector<int> push_right(int i, int n) {
  vector<int> v;
  while (i + 2 <= n && a[i] == 0 && a[i + 1] == 1 && a[i + 2] == 1) {
    op(i);
    v.push_back(i);
    i += 2;
  }
  return v;
}
int main() {
  int i, t;
  for (scanf("%d", &t); t--;) {
    int n;
    vector<int> v, ans;
    int cum = 0;
    for (scanf("%d", &n), i = 1; i <= n; ++i) {
      scanf("%d", a + i);
      cum += a[i];
    }
    if (cum == 0) {
      printf("YES\n0\n");
      continue;
    } else if ((cum & 1) || cum == n) {
      printf("NO\n");
      continue;
    }
    int cur = 1;
    if (a[1]) {
      cum = a[1];
      cur = 2;
      while (cur <= n && a[cur]) cum += a[cur], ++cur;
      if (cum & 1) {
        --cur;
        bool y = 1;
        while (cur <= n && a[cur]) {
          if (cur + 2 <= n) {
            if (a[cur] + a[cur + 1] + a[cur + 2] == 3) {
              cur += 2;
            } else if (a[cur + 1] + a[cur + 2] == 0) {
              ans.push_back(cur);
              op(cur);
              cur += 2;
            } else if (a[cur + 1] == 0 && a[cur + 2] == 1) {
              ans.push_back(cur);
              op(cur);
              auto push_vec = push_left(cur);
              ans.insert(ans.end(), push_vec.begin(), push_vec.end());
              cur += 3;
              break;
            } else {
              auto push_vec = push_left(cur + 2);
              ans.insert(ans.end(), push_vec.begin(), push_vec.end());
              cur += 3;
              break;
            }
          } else {
            y = 0;
            break;
          }
        }
        if (y == 0) {
          printf("NO\n");
          continue;
        }
      } else {
        ans = push_left(cur);
      }
    }
    cur = 1;
    while (cur + 2 <= n) {
      if (a[cur] == 0 && a[cur + 1] + a[cur + 2] == 2) {
        op(cur);
        ans.push_back(cur);
        cur += 2;
      } else if (a[cur]) {
        int st = cur - 1;
        while (1) {
          if (cur + 2 <= n && a[cur + 1] + a[cur + 2] == 2)
            cur += 2;
          else if (cur + 2 <= n && a[cur + 1] + a[cur + 2] == 0) {
            op(cur);
            ans.push_back(cur);
            cur += 2;
          } else if (cur + 2 <= n && a[cur + 1] == 0 && a[cur + 2] == 1) {
            op(cur);
            ans.push_back(cur);
            break;
          } else {
            break;
          }
        }
        auto push_vec = push_right(st, n);
        ans.insert(ans.end(), push_vec.begin(), push_vec.end());
      } else
        ++cur;
    }
    cum = 0;
    for (i = 1; i <= n; ++i) cum += a[i];
    if (cum == 0) {
      printf("YES\n%d\n", ans.size());
      for (auto x : ans) printf("%d ", x);
      puts("");
    } else
      printf("NO\n");
  }
  return 0;
}

```