#include<stdio.h>
#include<math.h>
int main()
{
int n1=0,n2=0,n3=0,n,cost,i=0;
scanf("%d",&n);
scanf("%d",&cost);
for(n1=0;n1<=n;n1++)
{
for(n2=0;n2<n-n1;n2++)//枚举女人人数,只能枚举到n-i,否则可能输出n2为负数的解
{
n3=n-n1-n2;//小孩人数
if(3*n1+2*n2+n3==cost)
{
i+=1;
printf("%d %d %d\n",n1,n2,n3);
}
}
}
if(i==0)
printf("No answer");
return 0;
}