/*

若干个 1~9，选出若干个数想加使得大于 5 的数尽量多
首先本来就大于等于 5 的数不加是最优的
接下来设 1 2 3 4 数量分别为 a b c d
1+4
2+3/4
3+4

a>d d then a-d
b>c c then b-c

*/

#include <bits/stdc++.h>

using namespace std;

inline int read()
{
	int x=0;bool f=0;char ch=getchar();
	while(ch<'0' || ch>'9') {if(ch=='-') f=1;ch=getchar();}
	while(ch>='0' && ch<='9') x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
	return f?-x:x;
}

#define x first
#define y second
#define pb push_back
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)

typedef pair<int,int> PII;
typedef long long LL;
typedef unsigned long long ull;
const int N=1e6+10;

int n;
int s[5];
float a[N];

int main()
{
	int T=read();
	while(T--) {
		n=read(),s[0]=s[1]=s[2]=s[3]=s[4]=0;
		int sum=0;
		rep(i,1,n) {
			scanf("%f",&a[i]),sum+=floor(a[i]);
			if(a[i]-floor(a[i])<0.5) s[(int)(10*(a[i]-floor(a[i])))]++;
			// cout<<(int)(10*(a[i]-floor(a[i])))<<a[i]-floor(a[i])<<endl;
		}
		int a=s[1],b=s[2],c=s[3],d=s[4],t1=min(a,d),t2=min(b,c);
		sum+=t1+t2;
		// cout<<d<<' '<<s[4]<<endl;
		a-=t1,d-=t1,b-=t2,c-=t2;
		// cout<<sum<<endl;
		if(a) {
			if(b) {
				if(b/2<=a) sum+=(b/2)+(a-b/2+2*(b&2))/5;
				else sum+=a+(b-a*2)*2/3;
			}
			else if(c) {
				if(a/2>=c) sum+=c+(a-c*2)/5;
				else sum+=(a/2)+((c-a/2)*3+(a&1))/5;
			}
			else sum+=a/5;
		}
		else if(d) sum+=(b+c+d)/2;
		// cout<<d<<' '<<t1<<endl;
		printf("%d\n",sum);
	}
	
	return 0;
}