codeforces981e(线段树+bitset)

E. Addition on Segments

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Grisha come to a contest and faced the following problem.

You are given an array of size $n$, initially consisting of zeros. The elements of the array are enumerated from $1$ to $n$. You perform $q$operations on the array. The $i$-th operation is described with three integers $l_i$, $r_i$ and $x_i$ ($1\le l_i\le r_i\le n$, $1\le x_i\le n$) and means that you should add $x_i$ to each of the elements with indices $l_i,l_i+1,\dots ,r_i$. After all operations you should find the maximum in the array.

Grisha is clever, so he solved the problem quickly.

However something went wrong inside his head and now he thinks of the following question: "consider we applied some subset of the operations to the array. What are the possible values of the maximum in the array?"

Help Grisha, find all integers $y$ between $1$ and $n$ such that if you apply some subset (possibly empty) of the operations, then the maximum in the array becomes equal to $y$.

Input

The first line contains two integers $n$ and $q$ ($1\le n,q\le {10}^4$) — the length of the array and the number of queries in the initial problem.

The following $q$ lines contain queries, one per line. The $i$-th of these lines contains three integers $l_i$, $r_i$ and $x_i$ ($1\le l_i\le r_i\le n$, $1\le x_i\le n$), denoting a query of adding $x_i$ to the segment from $l_i$-th to $r_i$-th elements of the array, inclusive.

Output

In the first line print the only integer $k$, denoting the number of integers from $1$ to $n$, inclusive, that can be equal to the maximum in the array after applying some subset (possibly empty) of the given operations.

In the next line print these $k$ integers from $1$ to $n$ — the possible values of the maximum. Print these integers in increasing order.

Examples

input

Copy

4 3
1 3 1
2 4 2
3 4 4

output

Copy

4
1 2 3 4 

input

Copy

7 2
1 5 1
3 7 2

output

Copy

3
1 2 3 

input

Copy

10 3
1 1 2
1 1 3
1 1 6

output

Copy

6
2 3 5 6 8 9 

Note

Consider the first example. If you consider the subset only of the first query, the maximum is equal to $1$. If you take only the second query, the maximum equals to $2$. If you take the first two queries, the maximum becomes $3$. If you take only the fourth query, the maximum becomes $4$. If you take the fourth query and something more, the maximum becomes greater that $n$, so you shouldn't print it.

In the second example you can take the first query to obtain $1$. You can take only the second query to obtain $2$. You can take all queries to obtain $3$.

In the third example you can obtain the following maximums:

• You can achieve the maximim of $2$ by using queries: $\left(1\right)$.
• You can achieve the maximim of $3$ by using queries: $\left(2\right)$.
• You can achieve the maximim of $5$ by using queries: $(1,2)$.
• You can achieve the maximim of $6$ by using queries: $\left(3\right)$.
• You can achieve the maximim of $8$ by using queries: $(1,3)$.
• You can achieve the maximim of $9$ by using queries: $(2,3)$.

题意：有一个长度为n的数组，标号为1到n，初始都为0，现在可以进行q次操作，每次操作把一个区间中的数都加上一个数，现在只做q次操作的子集，求问完成后数组中最大数可能是多少，把所有可能的都列出来。

思路：用线段树存储一段区间内可以进行的不同操作，然后利用STL库中的bitset，自顶向下搜寻，记录下区间可以达到的所有值，最后汇总。

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<bitset>
#include<vector>
using namespace std;
const int maxn = 1e4+10;
bitset<maxn> ans,y;
vector<int> tree[maxn << 2];
void update(int l,int r,int rt,int x,int y,int w)
{
    if(x>=l&&y<=r)
    {
        tree[rt].push_back(w);
        return ;
    }
    if(x == y)return;
    int mid = (x+y)>>1;
    update(l,r,rt<<1,x,mid,w);
    update(l,r,(rt<<1)+1,mid+1,y,w);
}
void dfs(int rt,bitset<maxn>y,int len)
{
    bitset<maxn>z=y;
    for(int i=tree[rt].size()-1;i>=0;i--)
    {
        z|=(z<<tree[rt][i]);
    }
    if(len == 1)
    {
        ans |= z;
        return ;
    }
    dfs(rt<<1,z,len-(len>>1));
    dfs((rt<<1)+1,z,len>>1);
}
int main()
{
    int n,q;
    while(~scanf("%d%d",&n,&q))
    {
        for(int i=0;i<=(n<<2);i++)tree[i].clear();
        ans = 0;
        y = 0;
        y[0]=1;
        int l,r,w;
        for(int i=0;i<q;i++)
        {
            scanf("%d%d%d",&l,&r,&w);
            update(l,r,1,1,n,w);
        }
        dfs(1,y,n);
        int cnt=0;
        for(int i=1;i<=n;i++)
        {
            if(ans[i])cnt++;
        }
        printf("%d\n",cnt);
        for(int i=1;i<=n;i++)
        {
            if(ans[i])
            {
                printf("%d%c",i,--cnt==0?'\n':' ');
            }
        }
    }
}