#include <bits/stdc++.h>
void sc(int &x) { scanf("%d", &x); }
void sc(int &x, int &y) { scanf("%d%d", &x, &y); }
void sc(int &x, int &y, int &z) { scanf("%d%d%d", &x, &y, &z); }
void sc(long long &x) { scanf("%lld", &x); }
void sc(long long &x, long long &y) { scanf("%lld%lld", &x, &y); }
void sc(long long &x, long long &y, long long &z) {
  scanf("%lld%lld%lld", &x, &y, &z);
}
void sc(char *x) { scanf("%s", x); }
void sc(char *x, char *y) { scanf("%s%s", x, y); }
void sc(char *x, char *y, char *z) { scanf("%s%s%s", x, y, z); }
void out(int x) { printf("%d\n", x); }
void out(long long x) { printf("%lld\n", x); }
void out(int x, int y) { printf("%d %d\n", x, y); }
void out(long long x, long long y) { printf("%lld %lld\n", x, y); }
void out(int x, int y, int z) { printf("%d %d %d\n", x, y, z); }
void out(long long x, long long y, long long z) {
  printf("%lld %lld %lld\n", x, y, z);
}
void ast(long long x, long long l, long long r) { assert(x >= l && x <= r); }
using namespace std;
const int N = 3e5 + 5, mod = 998244353;
int n, a[N];
int sum(int x) { return a[x] + a[x + 1] + a[x + 2]; }
void sol(int cas) {
  sc(n);
  for (int i = 1; i <= n; i++) sc(a[i]);
  vector<int> ans;
  int sm = 0, s = 0;
  for (int i = 1; i <= n; i++) sm += a[i];
  if (sm & 1) {
    printf("NO\n");
    return;
  }
  for (int i = 1; i <= n; i++)
    if (a[i] == 1) {
      int j = i;
      bool flag = false;
      while (j + 2 <= n) {
        if (sum(j) == 1) {
          ans.push_back(j);
          a[j + 1] = a[j + 2] = 1;
          j += 2;
        } else if (sum(j) == 2) {
          flag = true;
          for (int k = j; k >= i; k--)
            if (sum(k) == 2) {
              ans.push_back(k);
              a[k] = a[k + 1] = a[k + 2] = 0;
            }
          break;
        } else
          j += 2;
      }
      if (!flag) {
        if (i > 1) {
          for (int k = i - 1; k <= j - 1 && k <= n - 2; k++)
            if (sum(k) == 2) {
              ans.push_back(k);
              a[k] = a[k + 1] = a[k + 2] = 0;
            }
        }
      }
    }
  for (int i = 1; i <= n; i++)
    if (a[i]) {
      printf("NO\n");
      return;
    }
  printf("YES\n");
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++)
    printf(i == ans.size() - 1 ? "%d\n" : "%d ", ans[i]);
}
int main() {
  srand(time(0));
  int t = 1, cas = 0;
  scanf("%d", &t);
  while (t--) {
    sol(++cas);
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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>
void sc(int &x) { scanf("%d", &x); }
void sc(int &x, int &y) { scanf("%d%d", &x, &y); }
void sc(int &x, int &y, int &z) { scanf("%d%d%d", &x, &y, &z); }
void sc(long long &x) { scanf("%lld", &x); }
void sc(long long &x, long long &y) { scanf("%lld%lld", &x, &y); }
void sc(long long &x, long long &y, long long &z) {
  scanf("%lld%lld%lld", &x, &y, &z);
}
void sc(char *x) { scanf("%s", x); }
void sc(char *x, char *y) { scanf("%s%s", x, y); }
void sc(char *x, char *y, char *z) { scanf("%s%s%s", x, y, z); }
void out(int x) { printf("%d\n", x); }
void out(long long x) { printf("%lld\n", x); }
void out(int x, int y) { printf("%d %d\n", x, y); }
void out(long long x, long long y) { printf("%lld %lld\n", x, y); }
void out(int x, int y, int z) { printf("%d %d %d\n", x, y, z); }
void out(long long x, long long y, long long z) {
  printf("%lld %lld %lld\n", x, y, z);
}
void ast(long long x, long long l, long long r) { assert(x >= l && x <= r); }
using namespace std;
const int N = 3e5 + 5, mod = 998244353;
int n, a[N];
int sum(int x) { return a[x] + a[x + 1] + a[x + 2]; }
void sol(int cas) {
  sc(n);
  for (int i = 1; i <= n; i++) sc(a[i]);
  vector<int> ans;
  int sm = 0, s = 0;
  for (int i = 1; i <= n; i++) sm += a[i];
  if (sm & 1) {
    printf("NO\n");
    return;
  }
  for (int i = 1; i <= n; i++)
    if (a[i] == 1) {
      int j = i;
      bool flag = false;
      while (j + 2 <= n) {
        if (sum(j) == 1) {
          ans.push_back(j);
          a[j + 1] = a[j + 2] = 1;
          j += 2;
        } else if (sum(j) == 2) {
          flag = true;
          for (int k = j; k >= i; k--)
            if (sum(k) == 2) {
              ans.push_back(k);
              a[k] = a[k + 1] = a[k + 2] = 0;
            }
          break;
        } else
          j += 2;
      }
      if (!flag) {
        if (i > 1) {
          for (int k = i - 1; k <= j - 1 && k <= n - 2; k++)
            if (sum(k) == 2) {
              ans.push_back(k);
              a[k] = a[k + 1] = a[k + 2] = 0;
            }
        }
      }
    }
  for (int i = 1; i <= n; i++)
    if (a[i]) {
      printf("NO\n");
      return;
    }
  printf("YES\n");
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++)
    printf(i == ans.size() - 1 ? "%d\n" : "%d ", ans[i]);
}
int main() {
  srand(time(0));
  int t = 1, cas = 0;
  scanf("%d", &t);
  while (t--) {
    sol(++cas);
  }
  return 0;
}

```