Problem Description
Lawson is a magic swordsman with k kinds of magic attributes v1,v2,v3,…,vk. Now Lawson is faced with n monsters and the i-th monster also has k kinds of defensive attributes ai,1,ai,2,ai,3,…,ai,k. If v1≥ai,1 and v2≥ai,2 and v3≥ai,3 and … and vk≥ai,k, Lawson can kill the i-th monster (each monster can be killed for at most one time) and get EXP from the battle, which means vj will increase bi,j for j=1,2,3,…,k.
Now we want to know how many monsters Lawson can kill at most and how much Lawson’s magic attributes can be maximized.

Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line has two integers n and k (1≤n≤105,1≤k≤5).
The second line has k non-negative integers (initial magic attributes) v1,v2,v3,…,vk.
For the next n lines, the i-th line contains 2k non-negative integers ai,1,ai,2,ai,3,…,ai,k,bi,1,bi,2,bi,3,…,bi,k.
It’s guaranteed that all input integers are no more than 109 and vj+∑i=1nbi,j≤109 for j=1,2,3,…,k.

It is guaranteed that the sum of all n ≤5×105.
The input data is very large so fast IO (like fread) is recommended.

Output
For each test case:
The first line has one integer which means the maximum number of monsters that can be killed by Lawson.
The second line has k integers v′1,v′2,v′3,…,v′k and the i-th integer means maximum of the i-th magic attibute.

Sample Input
1
4 3
7 1 1
5 5 2 6 3 1
24 1 1 1 2 1
0 4 1 5 1 1
6 0 1 5 3 1

Sample Output
3
23 8 4

Hint

For the sample, initial V = [7, 1, 1]
① kill monster #4 (6, 0, 1), V + [5, 3, 1] = [12, 4, 2]
② kill monster #3 (0, 4, 1), V + [5, 1, 1] = [17, 5, 3]
③ kill monster #1 (5, 5, 2), V + [6, 3, 1] = [23, 8, 4]
After three battles, Lawson are still not able to kill monster #2 (24, 1, 1)
because 23 < 24.

看了题解之后发现不是一道很难的题目，主要是输入的问题，普通的read()已经满足不了它了。
我们可以开优先队列或者set来存满足条件的数，一旦i满足，我们就把i+1存到下一个队列里面，这样一次一次操作就可以吧所有可能达到的都放到最后一个队列里，如果没有新的值进来就说明已经到循环尾声，break；

#include<bits/stdc++.h>
using namespace std;

namespace fastIO {
    #define BUF_SIZE 100000
    //fread -> read
    bool IOerror = 0;
    inline char nc() {
        static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
        if(p1 == pend) {
            p1 = buf;
            pend = buf + fread(buf, 1, BUF_SIZE, stdin);
            if(pend == p1) {
                IOerror = 1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
    }
    inline void read(int &x) {
        char ch;
        while(blank(ch = nc()));
        if(IOerror) return;
        for(x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');
    }
    #undef BUF_SIZE
};
using namespace fastIO;

#define pa pair<int,int>
#define mp(a,b) make_pair(a,b)
priority_queue<pa,vector<pa>, greater<pa> >Q[15];
int def[100005][15],add[100005][15];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,k;
        int s[15];
        read(n),read(k);
        priority_queue<pa,vector<pa>, greater<pa> >Q[15];
        for(int i=1;i<=k;i++)
            read(s[i]);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=k;j++)
                read(def[i][j]);
            Q[1].push({def[i][1],i});
            for(int j=1;j<=k;j++)
                read(add[i][j]);
        }
        int ans=0;
        while(1)
        {
            int lastans=ans;
            for(int i=1;i<k;i++)
            {
                while(!Q[i].empty()&&Q[i].top().first<=s[i])
                {
                    int pos=Q[i].top().second;
                    Q[i].pop();
                    Q[i+1].push({def[pos][i+1],pos});

                }
            }
            while(!Q[k].empty()&&Q[k].top().first<=s[k])
            {
                int pos=Q[k].top().second;
                Q[k].pop();
                ans++;
                for(int i=1;i<=k;i++)
                    s[i]+=add[pos][i];
            }
            if(ans==lastans)
                break;
        }
        printf("%d\n",ans);
        for(int i=1;i<=k;i++)
            printf("%d%c",s[i],i==k?'\n':' ');
    }
    return 0;
}