Codeforces #381 C. Alyona and mex

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 31 32 54 5

output

21 0 2 1 0

input

4 21 42 4

output

35 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.

题目大意：  重要是理解mex的意思，我做的时候没有找到mex 的解释，一直懵逼，后来第二天才发现里面有一句，意思是：mex是这串子集中没有出现过的非负整数的最小值，现在n个元素的未知数组，给你提供了m个区间li到ri，每个区间找到一个mex值，然后比较，找出这么多区间中最小的那个mex值，现在想让我们尽可能使最终得到的mex值尽可能的大，让我们输出最后得出尽可能大的mex的值，然后再随便输出任意一组正确的数组。

题目分析：这道题读懂之后，首先要解决的是最终得到的最大的mex值，因为给定的区间，所以我们可以通过给定的区间进行判断，因为我们要使得到的最小值最大化，所以我们应该找到最小的区间长度，而这个区间长度就是最大的mex值。接下来是任意输出一个满足条件的数组。这个数组的构造刚开始想不到，但是后来想想，就是用i从0到n去mod x(mex)的值，0 1 ...x-1 0 1 ...x-1。即使是最小的区间，也肯定能被这个从0到x-1覆盖，因为本来就是基于这个最小区间才得到的值。

#include <cstdio>#include <cstring>#include <map>#include <algorithm>using namespace std;const int maxn = 100005;int a[maxn],b[maxn];int main(){int n, m;while(~(scanf("%d%d",&n,&m))){int minm=100000;int u,v;while(m--){scanf("%d%d",&u,&v);minm=min(abs(u-v)+1,minm);}printf("%d\n",minm);for(int i=0;i<n-1;i++){printf("%d ",i%minm);} printf("%d\n",(n-1)%minm);}return 0;}