Codeforces 740C. Alyona and mex 思路模拟

C. Alyona and mex

time limit per test：

2 seconds

memory limit per test：

256 megabytes

input：

standard input

output：

standard output

Alyona's mother wants to present an array of n non-negative integers to Alyona. The array should be special.

Alyona is a capricious girl so after she gets the array, she inspects m of its subarrays. Subarray is a set of some subsequent elements of the array. The i-th subarray is described with two integers li and ri, and its elements are a[li], a[li + 1], ..., a[ri].

Alyona is going to find mex for each of the chosen subarrays. Among these m mexes the girl is going to find the smallest. She wants this minimum mex to be as large as possible.

You are to find an array a of n elements so that the minimum mex among those chosen by Alyona subarrays is as large as possible.

The mex of a set S is a minimum possible non-negative integer that is not in S.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105).

The next m lines contain information about the subarrays chosen by Alyona. The i-th of these lines contains two integers li and ri(1 ≤ li ≤ ri ≤ n), that describe the subarray a[li], a[li + 1], ..., a[ri].

Output

In the first line print single integer — the maximum possible minimum mex.

In the second line print n integers — the array a. All the elements in a should be between 0 and 109.

It is guaranteed that there is an optimal answer in which all the elements in a are between 0 and 109.

If there are multiple solutions, print any of them.

Examples

input

5 3
1 3
2 5
4 5

output

2
1 0 2 1 0

input

4 2
1 4
2 4

output

3
5 2 0 1

Note

The first example: the mex of the subarray (1, 3) is equal to 3, the mex of the subarray (2, 5) is equal to 3, the mex of the subarray(4, 5) is equal to 2 as well, thus the minumal mex among the subarrays chosen by Alyona is equal to 2.

题目链接：http://codeforces.com/contest/740/problem/C

---

题意：定义mex为一个集合里面没有出现的最小的非负整数，有一个长度为n的集合，m个区间，现在要使得m各区间的最小的mex最大。

思路：最小的mex就是最小的区间长度。因为每个区间0,1,2,3...，才能使得最小的mex最大。当时最小的区间限制了长度，所以最小的mex最大为最小的区间长度x。长度为n的数组只要0,1,2,3,...x-2,x-1循环至n个数。因为每个区间的都是大于等于x的，所以无论怎么取区间都会有0~x-1。

代码：

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 int main()
 {
     int n,m;
     scanf("%d%d",&n,&m);
     int Max=1e6+;
     for(int i=; i<m; i++)
     {
         int l,r;
         scanf("%d%d",&l,&r);
         Max=min(Max,r-l+);
     }
     cout<<Max<<endl;
     int sign=;
     for(int i=; i<n; i++)
     {
         cout<<sign<<" ";
         sign++;
         sign=sign%Max;
     }
     cout<<endl;
     return ;
 }