B. Avoid Local Maximums
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given an array a of size n. Each element in this array is an integer between 1 and 109.

You can perform several operations to this array. During an operation, you can replace an element in the array with any integer between 1 and 109.

Output the minimum number of operations needed such that the resulting array doesn't contain any local maximums, and the resulting array after the operations.

An element ai is a local maximum if it is strictly larger than both of its neighbors (that is, ai>ai−1 and ai>ai+1). Since a1 and an have only one neighbor each, they will never be a local maximum.

Input

Each test contains multiple test cases. The first line will contain a single integer t (1≤t≤10000) — the number of test cases. Then t test cases follow.

The first line of each test case contains a single integer n (2≤n≤2⋅105) — the size of the array a.

The second line of each test case contains n integers a1,a2,…,an (1≤ai≤109), the elements of array.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105.

Output

For each test case, first output a line containing a single integer m — minimum number of operations required. Then ouput a line consist of n integers — the resulting array after the operations. Note that this array should differ in exactly m elements from the initial array.

If there are multiple answers, print any.

Example
input
Copy
5
3
2 1 2
4
1 2 3 1
5
1 2 1 2 1
9
1 2 1 3 2 3 1 2 1
9
2 1 3 1 3 1 3 1 3
output
Copy
0
2 1 2
1
1 3 3 1
1
1 2 2 2 1
2
1 2 3 3 2 3 3 2 1
2
2 1 3 3 3 1 1 1 3
Note

In the first example, the array contains no local maximum, so we don't need to perform operations.

In the second example, we can change a2 to 3, then the array don't have local maximums.

Solution



#include <bits/stdc++.h>

using namespace std;

#define int            long long int
#define F              first
#define S              second
#define pb             push_back
#define si             set <int>
#define vi             vector <int>
#define pii            pair <int, int>
#define vpi            vector <pii>
#define vpp            vector <pair<int, pii>>
#define mii            map <int, int>
#define mpi            map <pii, int>
#define spi            set <pii>
#define endl           "\n"
#define sz(x)          ((int) x.size())
#define all(p)         p.begin(), p.end()
#define double         long double
#define que_max        priority_queue <int>
#define que_min        priority_queue <int, vi, greater<int>>
#define bug(...)       __f (#__VA_ARGS__, __VA_ARGS__)
#define print(a)       for(auto x : a) cout << x << " "; cout << endl
#define print1(a)      for(auto x : a) cout << x.F << " " << x.S << endl
#define print2(a,x,y)  for(int i = x; i < y; i++) cout<< a[i]<< " "; cout << endl

inline int power(int a, int b)
{
  int x = 1;
  while (b)
  {
    if (b & 1) x *= a;
    a *= a;
    b >>= 1;
  }
  return x;
}

template <typename Arg1>
void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }
template <typename Arg1, typename... Args>
void __f (const char* names, Arg1&& arg1, Args&&... args)
{
  const char* comma = strchr (names + 1, ',');
  cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);
}

const int N = 200005;

void solve() {

int n; cin>>n;

vi v(n+1);
int ans=0;
for(int i=0; i<n; i++){
    cin>>v[i];
}
if(n<=2){
    cout<<0<<endl;
    for(int i=0; i<n; i++)cout<<v[i]<<" ";
    cout<<endl;
}
else{
    for(int i=1; i<n-1; i++){
        if(v[i]>v[i+1] && v[i]>v[i-1]){
            v[i+1]=max(v[i],v[i+2]);
            ans++;
        }
    }
    cout<<ans<<endl;
    for(int i=0; i<n; i++)cout<<v[i]<<" ";
    cout<<endl;
}



}

int32_t main()
{
  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);


  clock_t z = clock();

  int t = 1;
  cin >> t;
  while (t--) solve();



  return 0;
}