#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector")
using namespace std;
const int maxn = (int)2e5 + 10;
int v[maxn];
int n;
vector<int> ans;
void make(int pos) {
  assert(pos <= n - 3);
  int x = v[pos] ^ v[pos + 1] ^ v[pos + 2];
  v[pos] = v[pos + 1] = v[pos + 2] = x;
}
void solve(int l, int r) {
  if (r - l + 1 < 3) {
    assert(v[l] == 0 && v[r] == 0);
    return;
  }
  if (v[l] == 0) {
    if (v[l + 1] == 0) {
    } else if (v[l + 2] == 1) {
      ans.push_back(l);
      make(l);
    } else {
      ans.push_back(l + 1);
      make(l + 1);
      if (v[l + 1] == 1) {
        ans.push_back(l);
        make(l);
      }
    }
    solve(l + 1, r);
  } else {
    assert(v[r] == 0);
    if (v[r - 1] == 0) {
    } else if (v[r - 2] == 1) {
      ans.push_back(r - 2);
      make(r - 2);
    } else {
      ans.push_back(r - 3);
      make(r - 3);
      if (v[r - 1] == 1) {
        ans.push_back(r - 2);
        make(r - 2);
      }
    }
    solve(l, r - 1);
  }
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    ans.clear();
    scanf("%d", &n);
    int s = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &v[i]);
      s ^= v[i];
    }
    if (s) {
      printf("NO\n");
      continue;
    }
    int num = 0;
    bool ok = false;
    for (int i = 0; i < n; i++) {
      if (v[i] == 1) {
        num ^= 1;
      } else {
        if (num == 0) {
          solve(0, i);
          solve(i, n - 1);
          ok = true;
          break;
        }
      }
    }
    if (!ok) {
      int last = 0;
      for (int i = 1; i < n; i++) {
        if (v[i] == 1) {
          if ((i - last) % 2 == 0) {
            for (int j = last; j + 2 < i - 1; j += 2) {
              ans.push_back(j);
              make(j);
            }
            for (int j = i; j - 2 >= last; j -= 2) {
              ans.push_back(j - 2);
              make(j - 2);
            }
            solve(0, last);
            solve(i, n - 1);
            ok = true;
            break;
          }
          last = i;
        }
      }
    }
    if (!ok) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
    printf("%d\n", (int)ans.size());
    assert((int)ans.size() <= n);
    for (int x : ans) {
      printf("%d ", x + 1);
    }
    for (int i = 0; i < n; i++) {
      assert(v[i] == 0);
    }
    printf("\n");
  }
}

### 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("Ofast,no-stack-protector")
using namespace std;
const int maxn = (int)2e5 + 10;
int v[maxn];
int n;
vector<int> ans;
void make(int pos) {
  assert(pos <= n - 3);
  int x = v[pos] ^ v[pos + 1] ^ v[pos + 2];
  v[pos] = v[pos + 1] = v[pos + 2] = x;
}
void solve(int l, int r) {
  if (r - l + 1 < 3) {
    assert(v[l] == 0 && v[r] == 0);
    return;
  }
  if (v[l] == 0) {
    if (v[l + 1] == 0) {
    } else if (v[l + 2] == 1) {
      ans.push_back(l);
      make(l);
    } else {
      ans.push_back(l + 1);
      make(l + 1);
      if (v[l + 1] == 1) {
        ans.push_back(l);
        make(l);
      }
    }
    solve(l + 1, r);
  } else {
    assert(v[r] == 0);
    if (v[r - 1] == 0) {
    } else if (v[r - 2] == 1) {
      ans.push_back(r - 2);
      make(r - 2);
    } else {
      ans.push_back(r - 3);
      make(r - 3);
      if (v[r - 1] == 1) {
        ans.push_back(r - 2);
        make(r - 2);
      }
    }
    solve(l, r - 1);
  }
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    ans.clear();
    scanf("%d", &n);
    int s = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &v[i]);
      s ^= v[i];
    }
    if (s) {
      printf("NO\n");
      continue;
    }
    int num = 0;
    bool ok = false;
    for (int i = 0; i < n; i++) {
      if (v[i] == 1) {
        num ^= 1;
      } else {
        if (num == 0) {
          solve(0, i);
          solve(i, n - 1);
          ok = true;
          break;
        }
      }
    }
    if (!ok) {
      int last = 0;
      for (int i = 1; i < n; i++) {
        if (v[i] == 1) {
          if ((i - last) % 2 == 0) {
            for (int j = last; j + 2 < i - 1; j += 2) {
              ans.push_back(j);
              make(j);
            }
            for (int j = i; j - 2 >= last; j -= 2) {
              ans.push_back(j - 2);
              make(j - 2);
            }
            solve(0, last);
            solve(i, n - 1);
            ok = true;
            break;
          }
          last = i;
        }
      }
    }
    if (!ok) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
    printf("%d\n", (int)ans.size());
    assert((int)ans.size() <= n);
    for (int x : ans) {
      printf("%d ", x + 1);
    }
    for (int i = 0; i < n; i++) {
      assert(v[i] == 0);
    }
    printf("\n");
  }
}

```