题解--You are given a set of nn segments on the axis Ox--个性的点（水）

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#x轴与区间 #数组

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本文介绍了一道算法题目，任务是找出给定区间集合外的所有整数点。通过输入区间的数量及区间范围，利用C++编程实现，输出所有不在这些区间内的整数点。

网址：https://cn.vjudge.net/contest/243309#problem/A

You are given a set of n segments on the axis Ox, each segment has integer endpoints between 11 and mm inclusive. Segments may intersect, overlap or even coincide with each other. Each segment is characterized by two integers lili and riri (1≤li≤ri≤m1≤li≤ri≤m) — coordinates of the left and of the right endpoints.

Consider all integer points between 11 and mm inclusive. Your task is to print all such points that don't belong to any segment. The point xx belongs to the segment [l;r][l;r]if and only if l≤x≤rl≤x≤r.

Input

The first line of the input contains two integers nn and mm (1≤n,m≤1001≤n,m≤100) — the number of segments and the upper bound for coordinates.

The next nn lines contain two integers each lili and riri (1≤li≤ri≤m1≤li≤ri≤m) — the endpoints of the ii-th segment. Segments may intersect, overlap or even coincide with each other. Note, it is possible that li=rili=ri, i.e. a segment can degenerate to a point.

Output

In the first line print one integer kk — the number of points that don't belong to any segment.

In the second line print exactly kk integers in any order — the points that don't belong to any segment. All points you print should be distinct.

If there are no such points at all, print a single integer 00 in the first line and either leave the second line empty or do not print it at all.

Examples

Input

Output

2
3 4

Input

1 7
1 7

Output

Note

In the first example the point 11 belongs to the second segment, the point 22 belongs to the first and the second segments and the point 55 belongs to the third segment. The points 33 and 44 do not belong to any segment.

In the second example all the points from 11 to 77 belong to the first segment.

要求：给几组x轴上的区间，求x轴哪些整数没有在区间集合中出现过

代码：

#include<iostream>
using namespace std;
int main()
{
    int n,m,num;
    int l,r,point[103]={0,0,0};
    cin>>n>>m;
    num=m;
    for(int i=1;i<=n;i++){
        cin>>l>>r;
        for(int j=l;j<=r;j++){
            if(point[j]==1)
                continue;
            else {
                point[j]=1;
                num--;
            }
        }
    }
     cout<<(int)'z'<<endl;
    if(num==0){
        cout<<"0"<<endl;
        return 0;
    }
    cout<<num<<endl;
    for(int i=1;i<=m;i++){
        if(point[i]==0)
            cout<<i<<" " ;
    }
    return 0;
}