Evelyn likes drawing very much. Today, she draws lots of rainbows on white paper of infinite size, each using a different color. Since there're too many rainbows now, she wonders, how many of them can be seen?

For simplicity, each rainbow L_i is represented as a non-vertical line specified by the equation: y=a_ix+b_i. A rainbow L_i can be seen if there exists some x-coordinate x₀at which, its y-coordinate is strictly greater than y-coordinates of any other rainbows: a_ix₀+b_i > a_jx₀+b_j for all j != i.

Now, your task is, given the set of rainbows drawn, figure out the number of rainbows that can be seen.

Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <= 60) which is the number of test cases. And it will be followed by T consecutive test cases.

There's a blank line before every case. In each test case, there will first be an integer n (1 <= n <= 5000), which is the number of rainbows. Then n consecutive real number pairs follow. Each pair contains two real numbers, a_i and b_i, representing rainbow L_i: y=a_ix+b_i. No two rainbows will be the same, that is to say, have the same a and b.

Results should be directed to standard output. The output of each test case should be a single integer, which is the number of rainbows that can be seen.

2

1
1 1

3
1 0
2 0
3 0

1
2

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

typedef long long ll;
int n, m;

const int maxn=5010;
const double eps=1e-9;

struct node{
	double a,b;
}pp[maxn],p[maxn],stk[maxn];

int check(node x1,node x2,node x3){
	double tmp1=(x3.b-x2.b)/(x2.a-x3.a);
	double tmp2=(x2.b-x1.b)/(x1.a-x2.a);
	//printf("%.2f %.2f\n",tmp1,tmp2);
	if(tmp2+eps>=tmp1) return 1;
	return 0;
}

int cmp(node x1,node x2){
	if(x1.a==x2.a) return x1.b>x2.b;
	return x1.a<x2.a;
}

int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		int n;
		scanf("%d",&n);
		for(int i=1;i<=n;i++){
			scanf("%lf %lf",&p[i].a,&p[i].b);
		}
		sort(p+1,p+n+1,cmp);
		int cnt=0;
		pp[++cnt]=p[1];
		for(int i=2;i<=n;i++){
			if(fabs(p[i].a-p[i-1].a)<eps)continue;
			pp[++cnt]=p[i];
		}
		int top=-1;
		for(int i=1;i<=cnt;i++){
			if(top<=0){
				stk[++top]=pp[i];
			}
			else{
				node c=pp[i];
				while(top>0){
					node a=stk[top-1];
					node b=stk[top];
					if(check(a,b,c))//x2<=x1
						top--;
					else
						break;
				}
				stk[++top]=pp[i];
			}
		}
		printf("%d\n",top+1);
	}
	return 0;
}