#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T>
void ckmn(T& a, T b) {
  a = min(a, b);
}
template <typename T>
void ckmx(T& a, T b) {
  a = max(a, b);
}
void rd(int& x) { scanf("%i", &x); }
void rd(long long& x) { scanf("%lld", &x); }
void rd(char* x) { scanf("%s", x); }
void rd(double& x) { scanf("%lf", &x); }
void rd(string& x) { scanf("%s", &x); }
template <typename T1, typename T2>
void rd(pair<T1, T2>& x) {
  rd(x.first);
  rd(x.second);
}
template <typename T>
void rd(vector<T>& x) {
  for (T& i : x) rd(i);
}
template <typename T, typename... A>
void rd(T& x, A&... args) {
  rd(x);
  rd(args...);
}
template <typename T>
void rd() {
  T x;
  rd(x);
  return x;
}
int ri() {
  int x;
  rd(x);
  return x;
}
template <typename T>
vector<T> rv(int n) {
  vector<T> x(n);
  rd(x);
  return x;
}
template <typename T>
void ra(T a[], int n, int st = 1) {
  for (int i = 0; i < n; ++i) rd(a[st + i]);
}
template <typename T1, typename T2>
void ra(T1 a[], T2 b[], int n, int st = 1) {
  for (int i = 0; i < n; ++i) rd(a[st + i]), rd(b[st + i]);
}
template <typename T1, typename T2, typename T3>
void ra(T1 a[], T2 b[], T3 c[], int n, int st = 1) {
  for (int i = 0; i < n; ++i) rd(a[st + i]), rd(b[st + i]), rd(c[st + i]);
}
void re(vector<int> E[], int m, bool dir = 0) {
  for (int i = 0, u, v; i < m; ++i) {
    rd(u, v);
    E[u].push_back(v);
    if (!dir) E[v].push_back(u);
  }
}
template <typename T>
void re(vector<pair<int, T>> E[], int m, bool dir = 0) {
  for (int i = 0, u, v; i < m; ++i) {
    T w;
    rd(u, v, w);
    E[u].push_back({v, w});
    if (!dir) E[v].push_back({u, w});
  }
}
int main() {
  for (int t = ri(); t--;) {
    int n = ri();
    vector<int> a(n + 1, 0);
    for (int i = 1; i <= n; i++) rd(a[i]);
    int cnt1 = 0;
    for (int i = 1; i <= n; i++) cnt1 += a[i] == 1;
    if (cnt1 == n || cnt1 % 2 == 1)
      printf("NO\n");
    else {
      vector<int> ans;
      int ed = n, was = 0;
      while (ed > 0 && a[ed] == 0) ed--;
      for (int i = 1, j; i <= ed; i = j) {
        for (j = i; j <= n && a[j] == 1; j++) was++;
        if (i == j) {
          j++;
          continue;
        }
        if (was % 2 == 1) {
          while (j + 1 <= ed && a[j] == 0 && a[j + 1] == 0) {
            ans.push_back(j - 1);
            a[j] = 1;
            a[j + 1] = 1;
            j += 2;
          }
        }
      }
      cnt1 = 0;
      for (int i = 1; i <= n; i++) cnt1 += a[i] == 1;
      if (cnt1 == n || cnt1 % 2 == 1)
        printf("NO\n");
      else {
        printf("YES\n");
        int before = 0;
        for (int i = 1; i <= n - 2; i++) {
          if (before % 2 == 0 && a[i] + a[i + 1] + a[i + 2] == 2) {
            ans.push_back(i);
            a[i] = 0;
            a[i + 1] = 0;
            a[i + 2] = 0;
            int j = i - 2;
            while (j > 0 && a[j] + a[j + 1] + a[j + 2] == 2) {
              ans.push_back(j);
              a[j] = 0;
              a[j + 1] = 0;
              a[j + 2] = 0;
              j -= 2;
            }
          } else if (a[i] == 1)
            before++;
        }
        printf("%i\n", ans.size());
        for (int i : ans) printf("%i ", i);
        printf("\n");
      }
    }
  }
  return 0;
}

### Prompt

Your task is to create 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>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T>
void ckmn(T& a, T b) {
  a = min(a, b);
}
template <typename T>
void ckmx(T& a, T b) {
  a = max(a, b);
}
void rd(int& x) { scanf("%i", &x); }
void rd(long long& x) { scanf("%lld", &x); }
void rd(char* x) { scanf("%s", x); }
void rd(double& x) { scanf("%lf", &x); }
void rd(string& x) { scanf("%s", &x); }
template <typename T1, typename T2>
void rd(pair<T1, T2>& x) {
  rd(x.first);
  rd(x.second);
}
template <typename T>
void rd(vector<T>& x) {
  for (T& i : x) rd(i);
}
template <typename T, typename... A>
void rd(T& x, A&... args) {
  rd(x);
  rd(args...);
}
template <typename T>
void rd() {
  T x;
  rd(x);
  return x;
}
int ri() {
  int x;
  rd(x);
  return x;
}
template <typename T>
vector<T> rv(int n) {
  vector<T> x(n);
  rd(x);
  return x;
}
template <typename T>
void ra(T a[], int n, int st = 1) {
  for (int i = 0; i < n; ++i) rd(a[st + i]);
}
template <typename T1, typename T2>
void ra(T1 a[], T2 b[], int n, int st = 1) {
  for (int i = 0; i < n; ++i) rd(a[st + i]), rd(b[st + i]);
}
template <typename T1, typename T2, typename T3>
void ra(T1 a[], T2 b[], T3 c[], int n, int st = 1) {
  for (int i = 0; i < n; ++i) rd(a[st + i]), rd(b[st + i]), rd(c[st + i]);
}
void re(vector<int> E[], int m, bool dir = 0) {
  for (int i = 0, u, v; i < m; ++i) {
    rd(u, v);
    E[u].push_back(v);
    if (!dir) E[v].push_back(u);
  }
}
template <typename T>
void re(vector<pair<int, T>> E[], int m, bool dir = 0) {
  for (int i = 0, u, v; i < m; ++i) {
    T w;
    rd(u, v, w);
    E[u].push_back({v, w});
    if (!dir) E[v].push_back({u, w});
  }
}
int main() {
  for (int t = ri(); t--;) {
    int n = ri();
    vector<int> a(n + 1, 0);
    for (int i = 1; i <= n; i++) rd(a[i]);
    int cnt1 = 0;
    for (int i = 1; i <= n; i++) cnt1 += a[i] == 1;
    if (cnt1 == n || cnt1 % 2 == 1)
      printf("NO\n");
    else {
      vector<int> ans;
      int ed = n, was = 0;
      while (ed > 0 && a[ed] == 0) ed--;
      for (int i = 1, j; i <= ed; i = j) {
        for (j = i; j <= n && a[j] == 1; j++) was++;
        if (i == j) {
          j++;
          continue;
        }
        if (was % 2 == 1) {
          while (j + 1 <= ed && a[j] == 0 && a[j + 1] == 0) {
            ans.push_back(j - 1);
            a[j] = 1;
            a[j + 1] = 1;
            j += 2;
          }
        }
      }
      cnt1 = 0;
      for (int i = 1; i <= n; i++) cnt1 += a[i] == 1;
      if (cnt1 == n || cnt1 % 2 == 1)
        printf("NO\n");
      else {
        printf("YES\n");
        int before = 0;
        for (int i = 1; i <= n - 2; i++) {
          if (before % 2 == 0 && a[i] + a[i + 1] + a[i + 2] == 2) {
            ans.push_back(i);
            a[i] = 0;
            a[i + 1] = 0;
            a[i + 2] = 0;
            int j = i - 2;
            while (j > 0 && a[j] + a[j + 1] + a[j + 2] == 2) {
              ans.push_back(j);
              a[j] = 0;
              a[j + 1] = 0;
              a[j + 2] = 0;
              j -= 2;
            }
          } else if (a[i] == 1)
            before++;
        }
        printf("%i\n", ans.size());
        for (int i : ans) printf("%i ", i);
        printf("\n");
      }
    }
  }
  return 0;
}

```