#include <bits/stdc++.h>
using namespace std;
const int N=260, N2=90000;

struct node {
    int from, to;
    double x, y;
} a[N], e[N2]; 

int cnt;
bool cmp(node a, node b) { //使用atan2（-pi～pi）
    return atan2(a.x, a.y) < atan2(b.x, b.y);
}
int f[N2];//dp就是一维的dp，设f[i] 表示当前为第 i 个点时最多能选择多少个
bool vis[N2];
int main() {
    int n;scanf("%d", &n);
    for (int i = 1; i <= n; i++)scanf("%lf %lf", &a[i].x, &a[i].y);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)if (i != j) //n ^ 2 建边
            e[++cnt] ={i,j,a[j].x - a[i].x,a[j].y - a[i].y};
    sort(e + 1, e + 1 + cnt, cmp);
    int ans = 0;
    for (int i = 1; i <= n; i++) {
        memset(f, 0xc0, sizeof f);
        memset(vis, false, sizeof vis);
        vis[i] = true;
        f[i] = 0;
        for (int j = 1; j <= cnt; j ++) //一个个枚举，保证单调顺序
            if (vis[e[j].from]) {
                f[e[j].to] = max(f[e[j].to], f[e[j].from] + 1); //可以选可以不选
                vis[e[j].to] = true;
            }
        ans = max(ans, f[i]);
    }
    printf("%d", ans);
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
const int N=260, N2=90000;

struct node {
	int from, to;
	double x, y;
} a[N], e[N2]; 

int cnt;
bool cmp(node a, node b) { //使用atan2（-pi～pi）
	return atan2(a.x, a.y) < atan2(b.x, b.y);
}
int f[N2];//dp就是一维的dp，设f[i] 表示当前为第 i 个点时最多能选择多少个
bool vis[N2];
int main() {
	int n;scanf("%d", &n);
	for (int i = 1; i <= n; i++)scanf("%lf %lf", &a[i].x, &a[i].y);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)if (i != j) //n ^ 2 建边
            e[++cnt] ={i,j,a[j].x - a[i].x,a[j].y - a[i].y};
	sort(e + 1, e + 1 + cnt, cmp);
    int ans = 0;
	for (int i = 1; i <= n; i++) {
		memset(f, 0xc0, sizeof f);
		memset(vis, False, sizeof vis);
		vis[i] = True;
		f[i] = 0;
		for (int j = 1; j <= cnt; j ++) //一个个枚举，保证单调顺序
			if (vis[e[j].from]) {
				f[e[j].to] = max(f[e[j].to], f[e[j].from] + 1); //可以选可以不选
				vis[e[j].to] = True;
			}
		ans = max(ans, f[i]);
	}
	printf("%d", ans);
	return 0;
}