B

Problem description

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 l_i and r_i, and its elements are a[l_i], a[l_i + 1], ..., a[r_i].

Alyona is going to find mex for each of the chosen subarrays. Among these m mexesthe 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 l_i and r_i (1 ≤ l_i ≤ r_i ≤ n), that describe the subarray a[l_i], a[l_i + 1], ..., a[r_i].

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.

解题思路：最小的mex其实就是查看m个区间中哪个区间含有的元素个数最少。这道题最明显的就是这一点，然后接下来的一行输出（即a数组中的每个元素）让我一头雾水，完全找不到思路QAQ，完全不懂它究竟是怎么输出的=_=||，将题目精读了两个小时后，终于有了眉目。原来是要我们构造一个（新的数组元素）序列，结合红色语句可知，这个序列即数组a中的所有元素都可以是不超过最小mex这个最大值，即a[i](1<=i<=n)=i%mex（0~mex-1）；这样才保证mex大于集合S中m个区间各自的mex个元素值，即验证了题目中的这句话：集合S的mex是不在S中的最小可能的非负整数。还有一点，题目已经说明了如果有多种情况，打印其中任何一个值，因此这种构造序列的方法应该是正确的。

AC代码：

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n,m,l,r,mex=1e5+1;
    cin>>n>>m;
    while(m--){
        cin>>l>>r;
        mex=min(mex,r-l+1);
    }
    cout<<mex<<endl;
    for(int i=1;i<=n;++i)
        cout<<i%mex<<(i==n?"
":" ");
    return 0;
}