The code basically counts the number of nodes with even order and number of nodes with odd order and if they both are even individually, then it prints yes and no otherwise .

```
#include <iostream>
using namespace std;
int main(){
int x,y,even,odd,flag;
int test;
cin >> test;
int a[100];
int n,m;
for(int i = 0 ; i < test; i++){
even = odd = 0;
flag = 0;
cin >> n >> m;
for(int k = 0 ;k<=n;k++){
a[k] = 0;
}
for(int j = 0 ; j < m ; j++){
cin >> x >> y;
a[x]++;
a[y]++;
}
for(int j = 1 ; j <= n ; j++){
if(a[j] == 0){
flag++;
}else if(a[j]%2 != 0){
odd++;
}else{
even++;
}
}
if(flag == 0 && even%2 == 0 && odd%2 == 0){
cout << "YES" << endl;
}else{
cout << "NO" << endl;
}
}
return 0 ;
```

}