### My issue

Here I was using a approach by printing “-1” if n is more than 2 times of k;

then according to the question half of the elements must be odd and other be even, i took half of the elements has 1 and the the other elements expect the last one as 0; and the last element was the difference of k and half of the elements which were 1;

### My code

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin>>t;
while(t--){
int n,k;
int a[n];
cin>>n>>k;
if(n>2*k){
cout<<"-1"<<endl;
}
else{
for(int i=0;i<n/2;i++){
a[i]=1;
}
for(int i=n/2;i<n-1;i++){
a[i]=0;
}
a[n-1]=k-(n/2);
for(int i=0;i<n;i++){
cout<<a[i]<<" ";
}
cout<<endl;
}
}
return 0;
}
```

Problem Link: SUMARRAY Problem - CodeChef