CodeForces 430A Points and Segments

Description

Iahub isn't well prepared on geometry problems, but he heard that this year there will be a lot of geometry problems on the IOI selection camp. Scared, Iahub locked himself in the basement and started thinking of new problems of this kind. One of them is the following.

Iahub wants to draw n distinct points andm segments on theOX axis. He can draw each point with either red or blue. The drawing is good if and only if the following requirement is met: for each segment[l_i, r_i] consider all the red points belong to it (r_i points), and all the blue points belong to it (b_i points); each segmenti should satisfy the inequality|r_i - b_i| ≤ 1.

Iahub thinks that point x belongs to segment[l, r], if inequalityl ≤ x ≤ r holds.

Iahub gives to you all coordinates of points and segments. Please, help him to find any good drawing.

Input

The first line of input contains two integers: n (1 ≤ n ≤ 100) andm (1 ≤ m ≤ 100). The next line containsn space-separated integers x₁, x₂, ..., x_n (0 ≤ x_i ≤ 100) — the coordinates of the points. The following m lines contain the descriptions of them segments. Each line contains two integersl_i andr_i (0 ≤ l_i ≤ r_i ≤ 100) — the borders of thei-th segment.

It's guaranteed that all the points are distinct.

Output

If there is no good drawing for a given test, output a single integer -1. Otherwise outputn integers, each integer must be 0 or 1. Thei-th number denotes the color of thei-th point (0 is red, and 1 is blue).

If there are multiple good drawings you can output any of them.

Sample Input

Input

3 33 7 141 56 1011 15

Output

0 0 0

Input

3 41 2 31 22 35 62 2

Output

1 0 1 

题目大意：在数轴上给定n个点，m条线段，对于所有线段i，若点j满足li<=j<=ri，则说：点j在线段i上。要求将所给的点红蓝染色，使得对于任意一条线段，在这条线段上的红色点数ri，蓝色点数bi，满足|ri-bi|<=1。

解析：无论线段如何，只要点按照“0101010……”这种方法，从左到右画即可。最后要按照输入时的顺序输出，线段什么的不用考虑。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 110;struct POINT{int x;int id;int v;}point[N];bool cmp1(POINT a,POINT b) {return a.x < b.x;}bool cmp2(POINT a,POINT b) {return a.id < b.id;}int main() {int n,m;while( scanf("%d%d",&n,&m) != EOF) {for(int i = 0; i < n; i++) {scanf("%d",&point[i].x);point[i].id = i;}int a,b;for(int i = 0; i < m; i++) {scanf("%d%d",&a,&b);}sort(point,point+n,cmp1);for(int i = 0; i < n; i++) {point[i].v = i%2;}sort(point,point+n,cmp2);for(int i = 0; i < n; i++) {printf("%d",point[i].v);if(i != n) {printf(" ");}}printf("\n");}return 0;}