Codeforces Round #501 (Div. 3) 1015A Points in Segments （前缀和）

A. Points in Segments

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a set of nn segments on the axis OxOx , 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

Copy

3 5
2 2
1 2
5 5

Output

Copy

2
3 4 

Input

Copy

1 7
1 7

Output

Copy

0

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.

题目大意：n个线段 [l, r], 然后1~m个点，输出哪些点不属于任何线段。

当然是个水题，时间复杂度O(n*m), 如果  n,m <= 10^7呢？

可以用前缀和O(n+m)解答：

给出  l,r  暴力想法直接 vis[l]~vis[r] 标记 1，但是有更好的想法。

将 vis[l]++,  vis[r+1]-- ,求前缀和的时候，虽然中间的没有标记1，但是前缀和往后延申就达到标记效果了。将vis[r+1]--, 传到r+1的时候就（-1） + 1 = 0,  也就是没有出现过了。

例如 ：  点1  2  3 4 5

给出线段 【2 4】， 【4，5】；可以看出答案为   1

点              0    1    2     3      4      5       6

[2,4]           0     0    1     0     0      -1      0

[4,5]           0     0    1     0     1       0      -1

前缀和       0     0    1     1     2       2       1（只有点1为0）

AC代码(原题解)：

 #include <bits/stdc++.h>

 using namespace std;

 int main() {
 #ifdef _DEBUG
 //    freopen("input.txt", "r", stdin);
 //    freopen("output.txt", "w", stdout);
 #endif

     int n, m;
     cin >> n >> m;
     vector<int> cnt(m + );
     for (int i = ; i < n; ++i) {
         int l, r;
         cin >> l >> r;
         ++cnt[l];
         --cnt[r + ];
     }
     for (int i = ; i <= m; ++i)
         cnt[i] += cnt[i - ];
     vector<int> ans;
     for (int i = ; i <= m; ++i) {
         if (cnt[i] == )
             ans.push_back(i);
     }

     cout << ans.size() << endl;
     for (auto it : ans) cout << it << " ";
     cout << endl;

     return ;
 }