Problem:
Score : 500 points

Problem Statement

We have a two-dimensional grid with H \times W squares. There are M targets to destroy in this grid - the position of the i-th target is \left(h_i, w_i \right).

Takahashi will choose one square in this grid, place a bomb there, and ignite it. The bomb will destroy all targets that are in the row or the column where the bomb is placed. It is possible to place the bomb at a square with a target.

Takahashi is trying to maximize the number of targets to destroy. Find the maximum number of targets that can be destroyed.

Constraints

All values in input are integers.

1 \leq H, W \leq 3 \times 10^5

1 \leq M \leq \min\left(H\times W, 3 \times 10^5\right)

1 \leq h_i \leq H

1 \leq w_i \leq W

\left(h_i, w_i\right) \neq \left(h_j, w_j\right) \left(i \neq j\right)

Input

Input is given from Standard Input in the following format:

H W M
h_1 w_1
\vdots
h_M w_M

Output

Print the answer.

Sample Input 1

2 3 3
2 2
1 1
1 3

Sample Output 1

3

We can destroy all the targets by placing the bomb at \left(1, 2\right).

Sample Input 2

3 3 4
3 3
3 1
1 1
1 2

Sample Output 2

3

Sample Input 3

5 5 10
2 5
4 3
2 3
5 5
2 2
5 4
5 3
5 1
3 5
1 4

Sample Output 3

6

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(0);cin.tie(NULL)
#define ll long long
#define fm_t  ll t;cin>>t;while(t--)
#define fr(i,a,b) for( ll i=a;i<b;i++)
#define fr1(i,n,a) for( ll i=n-1;i>=a;i--)
#define sz(x) (ll)(x).size()
#define all(v)  (v).begin(), (v).end()
#define Sort(x) sort(x.begin(),x.end())
#define endl '\n'
#define pb push_back
#define ff first
#define ss second
const ll mod = 998244353;
#define Mod 1000000007

#define inf 1e18
#define ld long double
#define pll pair<ll, ll> 
#define vi vector<int>
#define vl vector<ll> 
#define vvl vector< vector<ll> > 
#define vvi vector< vector<int> >
#define vlp vector< pair<ll, ll> >
#define vllp vector<pair<ll, ll>, ll >
#define pi pair<ll, ll>
#define ppi pair<pair<ll, ll>, ll>
#define ump unordered_map<ll,ll>
#define pri priority_queue< pair<ll,ll> , vector< pair<ll,ll> >, greater< pair<ll,ll> > >
void ingraph(vl graph[], ll m){ll x, y;fr(i,0,m){cin>>x>>y;x--, y--;graph[x].pb(y);graph[y].pb(x);}}

#define max3(a, b, c)   max(max(a, b), c)
#define min3(a, b, c)   min(min(a, b), c)

//setprecision(10)
//greater<int>()
ll gcd(ll a, ll b){if(b==0)return a;return gcd(b, a%b);}
ll lcm(ll a, ll b){return a*b/gcd(a, b);}


ll modmulti(ll a, ll b){
   return ((a%mod)*1ll*(b%mod))%mod;
}
ll modadd(ll a, ll b){
   ll asdfgh =  ((a%mod)+(b%mod)+mod+mod)%mod;
   asdfgh = (asdfgh+mod)%mod;
   return asdfgh;
}
ll modpower(ll a, ll n){
   if(n==0) return 1;
   if(n==1) return a%mod;
   ll b = modpower(a,n/2);
   b = modmulti(b,b);
   if(n%2==0) return b;
   return modmulti(a,b);
}
ll modinv(ll a){
    return modpower(a,mod-2);
}   
     
int main()
{
      
  ll h,w,m;
  cin>>h>>w>>m;
  map<ll,ll>r,c;
  map<pair<ll,ll>,ll>mp;

  ll mx1=0,mx2=0;
  fr(i,0,m)
  {
    ll a,b;
    cin>>a>>b;
    r[a]++;
    c[b]++;
    mp[{a,b}]=1;
    
  }
ll ans=0;
  for(int i=1;i<=h;i++)
  {
    for(int j=1;j<=w;j++)
    {
      ans=max(ans,r[i]+c[j]-mp[{i,j}]);
    }
  }
  //cout<<c[1]<<endl;
  cout<<ans<<endl;


//cerr << "Time elapsed: " << clock() / (double)CLOCKS_PER_SEC << endl;
return 0;
}