Problem: Problem Statement

"Rooks Game" is a single-player game and uses a chessboard which has $N \times N$ grid and $M$ rook pieces.

A rook moves through any number of unoccupied squares horizontally or vertically. When a rook can attack another rook, it can capture the rook and move to the square which was occupied. Note that, in Rooks Game, we don't distinguish between white and black, in other words, every rook can capture any of other rooks.

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Initially, there are $M$ rooks on the board. In each move, a rook captures another rook. The player repeats captures until any rook cannot be captured. There are two types of goal of this game. One is to minimize the number of captured rooks, and the other is to maximize it.

In this problem, you are requested to calculate the minimum and maximum values of the number of captured rooks.

* * *

Input

The input consists of a single test case in the format below.

> $N$ $M$ $x_{1}$ $y_{1}$ $\vdots$ $x_{M}$ $y_{M}$

The first line contains two integers $N$ and $M$ which are the size of the chessboard and the number of rooks, respectively ($1 \le N, M \le 1000$). Each of the following $M$ lines gives the position of each rook. The $i$-th line with $x_{i}$ and $y_{i}$ means that the $i$-th rook is in the $x_{i}$-th column and $y_{i}$-th row ($1 \le x_{i}, y_{i} \le N$). You can assume any two rooks are not in the same place.

Output

Output the minimum and maximum values of the number of captured rooks separated by a single space.

Examples

Input| Output
---|---


8 3
1 1
1 8
8 8


|


1 2



8 4
1 1
1 8
8 8
8 1


|


2 3



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


|


0 0



100 1
100 100


|


0 0



10 12
1 2
1 4
1 6
3 6
5 6
10 7
8 7
6 7
4 7
2 7
7 3
9 5


|


7 8






Example

Input




Output

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define fi first
#define se second
#define inf (long long)1e8
using namespace std;
using p = pair<long long,long long>;

struct Unionfind{
    vector<int> par;
    vector<int> treerank;
    Unionfind(int n = 1){stree(n+1);}
    void stree(int n = 1){
        par.assign(n,-1);
        treerank.assign(n,0);
    }
    int root(int x){
        if(par[x] < 0)return x;
        return par[x] = root(par[x]);
    }
    bool issame(int x,int y){return root(x) == root(y);}
    bool uni(int x,int y){
        x = root(x);
        y = root(y);
        if(x == y)return 0;
        if(treerank[x] > treerank[y])swap(x,y);
        if(treerank[x] == treerank[y])++treerank[y];
        par[y] -= size(x);
        par[x] = y;
        return 1;
    }
    int size(int x){return -par[root(x)];}
};

template<class T>
struct Dinic{
    struct edge{
        int to;
        T cap;
        int rev;
    };
    vector<vector<edge>> lst;
    vector<int> completed;
    vector<int> dist;
    Dinic(int n = 1){
        lst.resize(n);
        completed.resize(n);
        dist.resize(n);
    }
    bool add(int start,int goal,T capacity){
        lst[start].push_back((edge){goal,capacity,(int)lst[goal].size()});
        lst[goal].push_back((edge{start,0,(int)lst[start].size()-1}));
        return 1;
    }
    void distbfs(int start){
        fill(dist.begin(),dist.end(),-1);
        queue<int> bfsqu;
        dist[start] = 0;
        bfsqu.push(start);
        while(bfsqu.size() > 0){
            int now = bfsqu.front();
            bfsqu.pop();
            for(int i=0;i < lst[now].size();++i){
                edge* nowe = &lst[now][i];
                if(nowe->cap > 0 && dist[nowe->to]<0){
                    dist[nowe->to] = dist[now] + 1;
                    bfsqu.push(nowe->to);
                }
            }
        }
    }
    T pathdfs(int now,int goal,T nf){
        if(now == goal)return nf;
        for(int& i= completed[now];i < lst[now].size();++i){
            edge* e = &lst[now][i];
            if(e->cap > 0 && dist[now] < dist[e->to]){
                T ans = pathdfs(e->to,goal,min(nf,e->cap));
                if(ans > 0){
                    e->cap -= ans;
                    lst[e->to][e->rev].cap += ans;
                    return ans;
                }
            }
        }
        return 0;
    }
    T solve(int start,int goal){
        T ans=0,nf = 0;
        while(1){
            distbfs(start);
            if(dist[goal]<0)return ans;
            fill(completed.begin(),completed.end(),0);
            while((nf=pathdfs(start,goal,inf))>0)ans += nf;
        }
        return -1;
    }
};

long long n,m;
long long dp[1005][1005][2][2] = {0};
vector<long long> lstx[1005],lsty[1005];
vector<p> v;
Unionfind uf;
set<p> st;

long long solve();

int main() {
    cin >> n >> m;
    v.resize(m);
    for (int i = 0; i < m; ++i){
        cin >> v[i].fi >> v[i].se;
        --v[i].fi;
        --v[i].se;
        st.insert(v[i]);
    }
    cout << solve() << endl;
    return 0;
}

long long solve(){
    long long ans = 0;
    uf = Unionfind(m);
    sort(v.begin(),v.end());
    for(int i = 0;i < m;++i){
        lstx[v[i].fi].push_back(i);
        lsty[v[i].se].push_back(i);
    }
    for(int i=0;i<n;++i){
        for(int j=1;j<lstx[i].size();++j)
            uf.uni(lstx[i][j],lstx[i][j-1]);
        for(int j=1;j<lsty[i].size();++j)
            uf.uni(lsty[i][j],lsty[i][j-1]);
    }
    //min
    Dinic<int> din(2 * n +2);
    for(int i = 0;i < m;++i)
        din.add(v[i].fi,v[i].se + n,1);
    for(int i = 0;i < n;++i){
        din.add(2 * n,i,1);
        din.add(i + n,2 * n + 1,1);
    }
    cout << m - din.solve(2 * n,2 * n+1)<< " ";
    // max
    ans = 0;
    vector<bool> ch(m,0);
    for(int i = 0;i < m;++i)
        if(!ch[uf.root(i)]) {
            ans += uf.size(uf.root(i)) - 1;
            ch[uf.root(i)] = 1;
        }
    return ans;
}