#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 2005;
int n;
int x[maxn];
int y[maxn];
int dp[maxn][maxn];	// dp[i][j]表示从i到j这一条边i可以收获的最大值
int f[maxn];		// f[i]表示从i点出发的可以收获的最大值
struct Edge{
	int u;//节点
	int v;//节点
	long long w;//价值
	Edge() {};//一定要有
	Edge(int a, int b){
		u = a;
		v = b;
		w = pow(x[a] - x[b], 2) + pow(y[a] - y[b], 2);//欧几里得距离的平方
	}
}e[maxn * maxn + maxn];
bool cmp(const Edge &x, const Edge &y){
	return x.w < y.w;
}
int total = 0;//记第几个
int main(){
	freopen("candy.in","r",stdin);
    freopen("candy.out","w",stdout);
    scanf("%d", &n);
	for (int i = 1; i <= n; i++){
		scanf("%d%d", x + i, y + i);
	}
	for (int i = 1; i <= n; i++){//加边
		e[total++] = Edge(0, i);
	}
	for (int i = 1; i <= n; i++){//加边
		for (int j = i + 1; j <= n; j++){
			e[total++] = Edge(i, j);
			e[total++] = Edge(j, i);
		}
	}
	sort(e, e + total, cmp);//排序
	for (int i = 0; i < total; i++){
		int j = i + 1;
		while (e[j].w == e[i].w){//由于题目中要求路径严格递减，所以长度相同的边必须特殊考虑，即同时更新。先更新边，再更新点。
			j++;
		}
		for (int k = i; k < j; k++){
			int &dis = dp[e[k].u][e[k].v];
			dis = max(dis, f[e[k].v] + 1);
		}
		for (int k = i; k < j; k++){
			Edge &tmp = e[k];
			f[tmp.u] = max(f[tmp.u], dp[tmp.u][tmp.v]);
		}
		i = j - 1;
	}
	printf("%d\n", f[0]);
    return 0;
}