#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2")


    #define int long long
    

#define YES cout<<"YES\n";return; 
#define NO cout<<"NO\n";return; 
#define write(a) cout<<a<<"\n";return; 
#define F first 
#define S second

using namespace std;
using vi=vector<int>; 
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count()); //uid<int> u1(5, 10); u1(rnd);
long long powmod(long long a,long long b,long long mod) {
        if (b==0 || a==1)  {if (mod==1) return 0; else return 1; }
       
        if (b%2==0) { long long k=powmod(a,b/2,mod); return (k*k)%mod; }
        else {long long k=powmod(a,b/2,mod); return ( (k*k) %mod *a)% mod; }
}
long long gcd(long long a, long long b) {
    if (a==0) return b;
    if (b==0) return a;
    if (a>b) return gcd(a%b,b); else return gcd(b%a,a);
}
int prime(int p) { // 1 - простое
if (p==1) return 0; 
    for (int i=2;i*i<=p;i++) {
    if (p%i==0 && i<p) return i;  
    }
    return 1;
}
        
 int inv(int a, int mod) {
    return powmod(a,mod-2,mod); 
}
            
 int random_modul() {
        
    uniform_int_distribution <int> u1(1e9, 2e9);
    int pepega=u1(rnd); 
    while (prime(pepega)!=1) pepega=u1(rnd); 
    return pepega; 
}
    
template <typename T1> void ifmax(T1 &max, T1 kek) {
    if (kek>max) max=kek; 
    return; 
}
template <typename T1> void ifmin(T1 &min, T1 kek) {
    if (kek<min) min=kek;
    return; 
}
     

template <typename T1> istream& operator>> (std::istream& in, pair<T1,T1> & a) {
in>>a.first>>a.second; 
return in; 
    
}
    
template <typename T1> istream& operator>> (std::istream& in, vector<T1> & a) {
    for (int i=0;i<a.size();i++) in>>a[i]; 
    return in;
} 
     
template <typename T1>  ostream& operator<< (std::ostream& out, vector <T1> &b) {
    for (int i=0;i<b.size();i++) out<<b[i]; 
    return out; 
} 
template <typename T1> void sort(vector <T1> &m) {
    sort(m.begin(),m.end());
}

template <typename T1> void reverse(vector <T1> &m) {
 reverse(m.begin(),m.end());
}
int mod=1000000007; 
int fac[200005];
int inv_fac[200005]; 
int co(int a, int b) {
    int ans=1;
    ans*=fac[b]; 
    ans*=inv_fac[b-a]; ans%=mod; 
    ans*=inv_fac[a]; 
    return ans%mod; 
}    


void solve () {
           
           
     /* ---- !! F, S !!  ---- */
         
   int a;
   cin>>a;
   vi m(a);
   cin>>m; 
   int s=0;
   for (int i=0;i<a;i++) {
       s+=m[i]; 
   }
   int u=0;
   int s1=0;
   int s2=0;
   for (int i=0;i<a;i++) {
       if (i%2==0) s1+=abs(m[i]-1); 
       else s2+=abs(m[i]-1); 
   }
   if (s1<=s2) {
       for (int i=0;i<a;i++) if (i%2==0) cout<<"1 "; else cout<<m[i]<<" "; 
       
       cout<<"\n"; return; 
   } else {
       for (int i=0;i<a;i++) if (i%2==1) cout<<"1 "; else cout<<m[i]<<" "; 
       
       cout<<"\n"; return; 
   }
  
    
    /* --------- */  
    return;
    }
         
         
         
    signed main() {
   ios_base::sync_with_stdio(0);
   cin.tie(0);
   cout.tie(0); 
           
           
      cout<<fixed<<setprecision(10); 
           
           
        int multitest = 1 ;    
           
      int tututu;
      tututu=1;
      
      
     int combinatorics = 0;  
      
      if (combinatorics) { 
     fac[1]=1;
     fac[0]=1; 
     for (int i=2;i<200005;i++) {
         fac[i]=fac[i-1]*i; fac[i]%=mod; 
     }
     for (int i=0;i<200005;i++) {
         inv_fac[i]=inv(fac[i],mod); 
     }
      } 
      
      
      
      if (multitest) cin>>tututu; // если нет запросов, то закоментить
            
      for (int qwerty=0;qwerty<tututu;qwerty++) solve();
          
           
    return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

You are given an array [a_1, a_2, ..., a_n] such that 1 ≤ a_i ≤ 10^9. Let S be the sum of all elements of the array a.

Let's call an array b of n integers beautiful if:

  * 1 ≤ b_i ≤ 10^9 for each i from 1 to n; 
  * for every pair of adjacent integers from the array (b_i, b_{i + 1}), either b_i divides b_{i + 1}, or b_{i + 1} divides b_i (or both); 
  * 2 ∑ _{i = 1}^{n} |a_i - b_i| ≤ S. 



Your task is to find any beautiful array. It can be shown that at least one beautiful array always exists.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each test case consists of two lines. The first line contains one integer n (2 ≤ n ≤ 50).

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).

Output

For each test case, print the beautiful array b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9) on a separate line. It can be shown that at least one beautiful array exists under these circumstances. If there are multiple answers, print any of them.

Example

Input


4
5
1 2 3 4 5
2
4 6
2
1 1000000000
6
3 4 8 1 2 3


Output


3 3 3 3 3
3 6
1 1000000000
4 4 8 1 3 3

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2")


    #define int long long
    

#define YES cout<<"YES\n";return; 
#define NO cout<<"NO\n";return; 
#define write(a) cout<<a<<"\n";return; 
#define F first 
#define S second

using namespace std;
using vi=vector<int>; 
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count()); //uid<int> u1(5, 10); u1(rnd);
long long powmod(long long a,long long b,long long mod) {
        if (b==0 || a==1)  {if (mod==1) return 0; else return 1; }
       
        if (b%2==0) { long long k=powmod(a,b/2,mod); return (k*k)%mod; }
        else {long long k=powmod(a,b/2,mod); return ( (k*k) %mod *a)% mod; }
}
long long gcd(long long a, long long b) {
    if (a==0) return b;
    if (b==0) return a;
    if (a>b) return gcd(a%b,b); else return gcd(b%a,a);
}
int prime(int p) { // 1 - простое
if (p==1) return 0; 
    for (int i=2;i*i<=p;i++) {
    if (p%i==0 && i<p) return i;  
    }
    return 1;
}
        
 int inv(int a, int mod) {
    return powmod(a,mod-2,mod); 
}
            
 int random_modul() {
        
    uniform_int_distribution <int> u1(1e9, 2e9);
    int pepega=u1(rnd); 
    while (prime(pepega)!=1) pepega=u1(rnd); 
    return pepega; 
}
    
template <typename T1> void ifmax(T1 &max, T1 kek) {
    if (kek>max) max=kek; 
    return; 
}
template <typename T1> void ifmin(T1 &min, T1 kek) {
    if (kek<min) min=kek;
    return; 
}
     

template <typename T1> istream& operator>> (std::istream& in, pair<T1,T1> & a) {
in>>a.first>>a.second; 
return in; 
    
}
    
template <typename T1> istream& operator>> (std::istream& in, vector<T1> & a) {
    for (int i=0;i<a.size();i++) in>>a[i]; 
    return in;
} 
     
template <typename T1>  ostream& operator<< (std::ostream& out, vector <T1> &b) {
    for (int i=0;i<b.size();i++) out<<b[i]; 
    return out; 
} 
template <typename T1> void sort(vector <T1> &m) {
    sort(m.begin(),m.end());
}

template <typename T1> void reverse(vector <T1> &m) {
 reverse(m.begin(),m.end());
}
int mod=1000000007; 
int fac[200005];
int inv_fac[200005]; 
int co(int a, int b) {
    int ans=1;
    ans*=fac[b]; 
    ans*=inv_fac[b-a]; ans%=mod; 
    ans*=inv_fac[a]; 
    return ans%mod; 
}    


void solve () {
           
           
     /* ---- !! F, S !!  ---- */
         
   int a;
   cin>>a;
   vi m(a);
   cin>>m; 
   int s=0;
   for (int i=0;i<a;i++) {
       s+=m[i]; 
   }
   int u=0;
   int s1=0;
   int s2=0;
   for (int i=0;i<a;i++) {
       if (i%2==0) s1+=abs(m[i]-1); 
       else s2+=abs(m[i]-1); 
   }
   if (s1<=s2) {
       for (int i=0;i<a;i++) if (i%2==0) cout<<"1 "; else cout<<m[i]<<" "; 
       
       cout<<"\n"; return; 
   } else {
       for (int i=0;i<a;i++) if (i%2==1) cout<<"1 "; else cout<<m[i]<<" "; 
       
       cout<<"\n"; return; 
   }
  
    
    /* --------- */  
    return;
    }
         
         
         
    signed main() {
   ios_base::sync_with_stdio(0);
   cin.tie(0);
   cout.tie(0); 
           
           
      cout<<fixed<<setprecision(10); 
           
           
        int multitest = 1 ;    
           
      int tututu;
      tututu=1;
      
      
     int combinatorics = 0;  
      
      if (combinatorics) { 
     fac[1]=1;
     fac[0]=1; 
     for (int i=2;i<200005;i++) {
         fac[i]=fac[i-1]*i; fac[i]%=mod; 
     }
     for (int i=0;i<200005;i++) {
         inv_fac[i]=inv(fac[i],mod); 
     }
      } 
      
      
      
      if (multitest) cin>>tututu; // если нет запросов, то закоментить
            
      for (int qwerty=0;qwerty<tututu;qwerty++) solve();
          
           
    return 0;
}

```