/*********************************
* 日期:2013-1-13
* 作者:SJF0115
* 题号: 九度1155
* 题目:鸡兔同笼
* 来源:http://ac.jobdu.com/problem.php?pid=1155
* 结果:AC
* 题意:
* 总结:
**********************************/
#include <stdio.h>
#include <string.h>
int main(){
int i,j,n,a,count,flag;
//freopen("C:\\Users\\SJF\\Desktop\\acm.txt","r",stdin);
while(scanf("%d",&n) != EOF){
//n组测试
for(i = 1;i <= n;i++){
scanf("%d",&a);
flag = 0;
int min = 32769,max = -1;
//2*x + 4*y = a
for(j = 0;j < 32769;j++){
if(4*j > a){
break;
}
else if((a - 4*j)%2 == 0){
flag = 1;
count = j+(a-4*j)/2;
if(min > count){
min = count;
}
if(max < count){
max = count;
}
}
}
if(flag == 0){
printf("0 0\n");
}
else{
printf("%d %d\n",min,max);
}
}
}
return 0;
}
/*********************************
* 日期:2013-1-15
* 作者:SJF0115
* 题号: 九度1155
* 题目:鸡兔同笼
* 来源:http://ac.jobdu.com/problem.php?pid=1155
* 结果:AC
* 题意:
* 总结:要使动物数最多应使鸡的数目尽可能的多;要使动物数最少应使鸡的数目尽可能的少。
**********************************/
#include <stdio.h>
#include <stdlib.h>
int main()
{
int n, i,a;
int max, min;
//freopen("C:\\Users\\SJF\\Desktop\\acm.txt","r",stdin);
while(scanf("%d",&n) != EOF)
{
for(i = 0; i < n; i ++)
{
scanf("%d", &a);
if(a % 2 != 0)
{
printf("0 0\n");
}
else
{
max = a / 2;
if(a % 4 != 0)
{
min = a / 4 + 1;
}
else
{
min = a / 4;
}
printf("%d %d\n",min, max);
}
}
}
return 0;
}