CodeForces - 1004B - Sonya and Exhibition（纯思维题）

题目链接：

http://codeforces.com/problemset/problem/1004/B

题目：

Sonya decided to organize an exhibition of flowers. Since the girl likes only roses and lilies, she decided that only these two kinds of flowers should be in this exhibition.

There are n flowers in a row in the exhibition. Sonya can put either a rose or a lily in the i-th position. Thus each of n positions should contain exactly one flower: a rose or a lily.

She knows that exactly m people will visit this exhibition. The i-th visitor will visit all flowers from li to ri inclusive. The girl knows that each segment has its own beauty that is equal to the product of the number of roses and the number of lilies.

Sonya wants her exhibition to be liked by a lot of people. That is why she wants to put the flowers in such way that the sum of beauties of all segments would be maximum possible.

Input

The first line contains two integers n and m (1≤n,m≤1e3) — the number of flowers and visitors respectively.

Each of the next m lines contains two integers li and ri (1≤li≤ri≤n), meaning that i-th visitor will visit all flowers from li to ri inclusive.

Output

Print the string of nn characters. The i-th symbol should be «0» if you want to put a rose in the i-th position, otherwise «1» if you want to put a lily.

If there are multiple answers, print any.

Examples

Input

5 3
1 3
2 4
2 5
      

Output

01100      

Input

6 3
5 6
1 4
4 6
      

Output

110010      

Note

In the first example, Sonya can put roses in the first, fourth, and fifth positions, and lilies in the second and third positions;

• in the segment [1…3], there are one rose and two lilies, so the beauty is equal to 1⋅2=2;
• in the segment [2…4], there are one rose and two lilies, so the beauty is equal to 1⋅2=2;
• in the segment [2…5], there are two roses and two lilies, so the beauty is equal to 2⋅2=4.

The total beauty is equal to 2+2+4=8.

In the second example, Sonya can put roses in the third, fourth, and sixth positions, and lilies in the first, second, and fifth positions;

• in the segment [5…6], there are one rose and one lily, so the beauty is equal to 1⋅1=1;
• in the segment [1…4], there are two roses and two lilies, so the beauty is equal to 2⋅2=4;
• in the segment [4…6], there are two roses and one lily, so the beauty is equal to 2⋅1=2.

The total beauty is equal to 1+4+2=7.

题目大意：

有一排n个格子，每个格子里能放一种花，一共有两种花，一种用 0 代表，另一种用 1 代表，然后给你m各区间，每个区间的价值就是这个区间内的两种花的数量之积。问你应该怎么放花，使得这些区间的价值和最大。

分析：

题目的意思转化一下，就是说让0 1 的个数在各个区间内都是接近的（和相等，越接近，积越大），也就是说0 1 分布均匀，那么，我们直接0 1 交替输出，就可以保证0 1 在各个区间都是最接近的。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	int l,r;
	for(int i=1;i<=m;i++)
		scanf("%d%d",&l,&r);
	for(int i=0;i<n;i++)
		printf("%d",i%2);
	putchar('\n');
	return 0;
}