CodeForces - 1004B

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

She knows that exactly mm people will visit this exhibition. The ii-th visitor will visit all flowers from lili to riri 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 nn and mm (1≤n,m≤1031≤n,m≤103) — the number of flowers and visitors respectively.

Each of the next mm lines contains two integers lili and riri (1≤li≤ri≤n1≤li≤ri≤n), meaning that ii-th visitor will visit all flowers from lili to riri inclusive.

Output

Print the string of nn characters. The ii-th symbol should be «0» if you want to put a rose in the ii-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][1…3], there are one rose and two lilies, so the beauty is equal to 1⋅2=21⋅2=2;
• in the segment [2…4][2…4], there are one rose and two lilies, so the beauty is equal to 1⋅2=21⋅2=2;
• in the segment [2…5][2…5], there are two roses and two lilies, so the beauty is equal to 2⋅2=42⋅2=4.

The total beauty is equal to 2+2+4=82+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][5…6], there are one rose and one lily, so the beauty is equal to 1⋅1=11⋅1=1;
• in the segment [1…4][1…4], there are two roses and two lilies, so the beauty is equal to 2⋅2=42⋅2=4;
• in the segment [4…6][4…6], there are two roses and one lily, so the beauty is equal to 2⋅1=22⋅1=2.

The total beauty is equal to 1+4+2=71+4+2=7.

 

　　一种新的思维方式，不是 根据查询要求去 构造满足查询区间要求的值 ，而是 构造一个完美的序列，使得它对任意的查询，都满足要求。 （不要为了活着而生活，因为生活所以活着）

 

　　题目要求的是求一个最大的数，区间中两种不同花的数量的乘积，不难发现，应当尽可能的使区间中两种花的数量都变大，不能让其中一种花的数目很大，而另一种很小，因为这样对一些查询区间是不合适的。为避免这种情况，所以应该是让两种花交替的种植。

 

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<iostream>
using namespace std;
typedef long long  LL;
const double pi=acos(-1.0);
const double e=exp(1);
const int N = 10;

int main()
{
    int n,m,t;
    int x,y,i,p,j;
    scanf("%d%d",&n,&m);
    for(i=1;i<=m;i++)
    {
        scanf("%d%d",&x,&y);
    }
    for(i=1;i<=n;i++)
    {
        if(i%2)
            printf("1");
        else
            printf("0");
    }
    putchar('\n');
    return 0;
}