Morgana is learning computer vision, and he likes cats, too. One day he wants to find the cat movement from a cat video. To do this, he extracts cat features in each frame. A cat feature is a two-dimension vector <xx, yy>. If x_ixi​= x_jxj​ and y_iyi​ = y_jyj​, then <x_ixi​, y_iyi​> <x_jxj​, y_jyj​> are same features.

So if cat features are moving, we can think the cat is moving. If feature <aa, bb> is appeared in continuous frames, it will form features movement. For example, feature <aa , bb > is appeared in frame 2,3,4,7,82,3,4,7,8, then it forms two features movement 2-3-42−3−4 and 7-87−8 .

Now given the features in each frames, the number of features may be different, Morgana wants to find the longest features movement.

Input

First line contains one integer T(1 \le T \le 10)T(1≤T≤10) , giving the test cases.

Then the first line of each cases contains one integer nn (number of frames),

In The next nn lines, each line contains one integer k_iki​ ( the number of features) and 2k_i2ki​ intergers describe k_iki​features in ith frame.(The first two integers describe the first feature, the 33rd and 44th integer describe the second feature, and so on).

In each test case the sum number of features NN will satisfy N \le 100000N≤100000 .

Output

For each cases, output one line with one integers represents the longest length of features movement.

样例输入复制

1

8

2 1 1 2 2

2 1 1 1 4

2 1 1 2 2

2 2 2 1 4

0

0

1 1 1

1 1 1

样例输出复制

3

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

题意：每一个frame中都有k个坐标

问连续出现的坐标 连续的frame最长的是多少

思路：

很简单的一个暴力 给每一个动作都编号 记下都在哪些frame中出现过

然后排个序 从头到尾跑一遍

要注意一个动作可能在同一个frame中出现

WA了超级久 初始化没有写好

看来以后还是不能在毛概课上A题

#include<iostream>

#include<stdio.h>

#include<string.h>

#include<algorithm>

#include<stack>

#include<queue>

#include<map>

#include<vector>

#include<set>

//#include<bits/stdc++.h>

#define inf 0x7f7f7f7f7f7f7f7f

using namespace std;

typedef long long LL;

const int maxn = 1e5+ 100;

typedef pair<int, int> mpair;

map<mpair, int> mmap;

int t, n, cntid, cnt;

struct node{

    int id, frame;

}feature[maxn];

bool cmp(node a, node b)

{

    if(a.id == b.id){

        return a.frame < b.frame;

    }

    else{

        return a.id < b.id;

    }

}

void init()

{

    mmap.clear();

    for(int i = 0; i < maxn; i++){

        feature[i].id = 0;

        feature[i].frame = 0;

    }

    cntid = 0;

    cnt = 0;

}

int main()

{

    cin>>t;

    while(t--){

        init();

        scanf("%d", &n);

        if(n == 0){

            printf("0\n");

            continue;

        }

        for(int i = 0; i < n; i++){

            int k;

            scanf("%d", &k);

            for(int j = 0; j < k; j++){

                int x, y;

                scanf("%d%d", &x, &y);

                mpair p = make_pair(x, y);

                if(mmap.find(p) == mmap.end()){

                    mmap[p] = cntid++;

                }

                feature[cnt].id = mmap[p];

                feature[cnt].frame = i;

                cnt++;

            }

        }

        sort(feature, feature + cnt, cmp);

        int ans = 1, tmp = 1;

        for(int i = 0; i < cnt - 1; i++){

            if(feature[i].id == feature[i + 1].id && feature[i].frame + 1 == feature[i + 1].frame){

                tmp++;

            }

            else if(feature[i].id == feature[i + 1].id && feature[i].frame == feature[i + 1].frame){

                continue;

            }

            else{

                ans = max(ans, tmp);

                tmp = 1;

            }

        }

        ans = max(ans, tmp);

        printf("%d\n", ans);

    }

	return 0;

}