Airdrop

Time Limit: 2 Seconds      Memory Limit: 65536 KB

PUBG is a multiplayer online battle royale video game. In the game, up to one hundred players parachute onto an island and scavenge for weapons and equipment to kill others while avoiding getting killed themselves. Airdrop in this game is a key element, as airdrops often carry with them strong weapons or numerous supplies, helping players to survive.

Airdrop in the game(?)

Consider the battle field of the game to be a two-dimensional plane. An airdrop has just landed at point  (both  and  are integers), and all the  players on the battle field, where  (both  and are integers) indicates the initial position of the -th player, start moving towards the airdrop with the following pattern:

• If the position of a living player at the beginning of this time unit is not equal to , he will begin his next move.
  • Let's say he is currently at point . For his next move, he will consider four points , ,  and .
  • He will select one of the four points whose Manhattan distance to the airdrop  is the smallest to be the destination of his next move. Recall that the Manhattan distance between two points  and  is defined as .
  • If two or more points whose Manhattan distance to the airdrop is the same, he will use the following priority rule to break the tie:  has the highest priority to be selected,  has the second highest priority,  has the third highest priority, and  has the lowest priority.
  • At the end of this time unit, he arrives at his destination.
• If the position of a living player at the beginning of this time unit is equal to , he will continue to fatten his backpack with the supplies in the airdrop and stays at .

But the battle is tough and it's almost impossible for all the players to arrive at the airdrop safely. If two or more players meet at point  other than , where both  and  are integers, they will fight and kill each other and none of them survive.

BaoBao is a big fan of the game and is interested in the number of players successfully arriving at the position of the airdrop, but he doesn't know the value of . Given the value of  and the initial position of each player, please help BaoBao calculate the minimum and maximum possible number of players successfully arriving at the position of the airdrop for all , where  is the set of all integers (note that  can be positive, zero or negative).

Input

There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:

The first line contains two integers  and  (, ), indicating the number of players and the  value of the airdrop.

For the following  lines, the -th line contains two integers  and  (), indicating the initial position of the -th player.

It's guaranteed that the sum of  in all test cases will not exceed , and in each test case no two players share the same initial position.

Output

For each test case output one line containing two integers  and  separated by one space.  indicates the minimum possible number of players successfully arriving at the position of the airdrop, while  indicates the maximum possible number.

Sample Input

3
3 2
1 2
2 1
3 5
3 3
2 1
2 5
4 3
2 3
1 3
4 3

Sample Output

1 3
0 3
2 2

Hint

We now explain the first sample test case.

To obtain the answer of , one should consider . The following table shows the position of each player at the end of each time unit when .

To obtain the answer of , one should consider . The following table shows the position of each player at the end of each time unit when .

解題思路：因為是固定Y座標，所以我們可以列舉X座標，X座標最多也就1e5而已。然後對於每個X座標，我們暴力的計算所有玩家到那個座標的距離，然後判斷即可。但是這麼做複雜度是N^2的，所以我們要高效的計算。我們列舉X座標時，計算他左邊能有多少個倖存者，和右邊有多少個倖存者即可，我們可以先從左向右列舉，然後可以實現快速轉移的計算X座標左邊有多少個生存者，然後從右往左列舉，計算右邊有多少個倖存者，結合答案即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN=200010;
int N,X,Y;
int x[MAXN];
int y[MAXN];

int num[MAXN*2];
int tot=0;
vector<int> G[MAXN];

int ans[MAXN];
int vis[MAXN];
vector<int> used;

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&N,&Y);
        tot=0;
        memset(vis,0,sizeof(vis));
        memset(ans,0,sizeof(ans));

        num[tot++]=0;
        for(int i=1;i<=N;i++){
            scanf("%d%d",&x[i],&y[i]);
            G[x[i]].clear();
            G[x[i]+1].clear();
            num[tot++]=x[i];
            num[tot++]=x[i]+1;
        }

        for(int i=1;i<=N;i++)
            G[x[i]].push_back(y[i]);


        sort(num,num+tot);
        tot = unique(num,num+tot)-num;


        int ansmin=0x3f3f3f3f;
        int ansmax=0;

        int pre=0;
        int cnt=0;
        for(int i=0;i<tot;i++){
            X=num[i];
            used.clear();
            ans[X]+=G[X].size();

            for(int j=0;j<G[pre].size();j++){
                int dis=abs(1e5+1-pre)+abs(Y-G[pre][j]);
                vis[dis]++;
                if(vis[dis]==1)
                    cnt++;
                else
                    used.push_back(dis);
            }
            for(int j=0;j<used.size();j++)
            {
                if(vis[used[j]])cnt--;
                vis[used[j]]=0;
            }
            ans[X]+=cnt;
            pre=num[i];
        }

        memset(vis,0,sizeof(vis));
        pre=cnt=0;
        for(int i=tot-1;i>=0;i--){
            X=num[i];
            used.clear();
            for(int j=0;j<G[pre].size();j++){
                int dis=abs(-pre)+abs(Y-G[pre][j]);
                vis[dis]++;
                if(vis[dis]==1)
                    cnt++;
                else
                    used.push_back(dis);
            }
            for(int j=0;j<used.size();j++)
            {
                if(vis[used[j]])cnt--;
                vis[used[j]]=0;
            }
            ans[X]+=cnt;
            ansmin=min(ans[X],ansmin);
            ansmax=max(ans[X],ansmax);
            pre=num[i];
        }
        printf("%d %d\n",ansmin,ansmax);
    }

    return 0;
}