A bipartite graph G = (V, E) is a graph in which the vertex set V can be divided into two disjoint subsets X and Y such that every edge e ∈ E has one end point in X and the other end point in Y.

A matching M is a subset of edges such that each node in V appears in at most one edge in M.

Given a bipartite graph, find the size of the matching which has the largest size.

Constraints

* 1 ≤ |X|, |Y| ≤ 100
* 0 ≤ |E| ≤ 10,000

Input


|X| |Y| |E|
x0 y0
x1 y1
:
x|E|-1 y|E|-1


|X| and |Y| are the number of vertices in X and Y respectively, and |E| is the number of edges in the graph G. The vertices in X are named with the numbers 0, 1,..., |X|-1, and vertices in Y are named with the numbers 0, 1,..., |Y|-1, respectively.

xi and yi are the node numbers from X and Y respectevely which represent the end-points of the i-th edge.

Output

Print the largest size of the matching.

Example

Input

3 4 6
0 0
0 2
0 3
1 1
2 1
2 3


Output

3
#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>

using namespace std;

#define INF 1e9+7

struct edge{
  int t,c, rev;
};

struct Graph{
  vector<vector<edge>> edges;
  vector<int> used;
  Graph(int n): edges(n){};
};

int dfs(int v, int t, int f, Graph& G){
  if(v == t) return f;
  G.used[v] = 1;
  for(edge& e:G.edges[v]){
    if(!G.used[e.t] && e.c > 0){
      int d = dfs(e.t, t, min(f, e.c), G);
      if(d > 0){
        e.c -= d;
        G.edges[e.t][e.rev].c += d;
        return d;
      }
    }
  }
  return 0;
}

int main(){
  int x,y,e;
  int ans = 0;
  cin >> x >> y >> e;
  Graph G = Graph(x+y+2);
  for(int i = 0; i < e; i++){
    int f,t;
    cin >> f >> t;
    G.edges[f].push_back((edge){x+t, 1, (int)G.edges[x+t].size()});
    G.edges[x+t].push_back((edge){f, 0, (int)G.edges[f].size() - 1});
  }
  int s = x+y;
  int t = x+y+1;

  for(int i = 0; i < x; i++){
    G.edges[s].push_back({i, 1, (int)G.edges[i].size()});
    G.edges[i].push_back({s, 0, (int)G.edges[s].size() - 1});
  }
  for(int i = 0; i < y; i++){
    G.edges[x+i].push_back({t, 1, (int)G.edges[t].size()});
    G.edges[t].push_back({x+i, 0, (int)G.edges[x+i].size() - 1});
  }
  for(;;){
    G.used = vector<int>(x+y+2, 0);
    int f = dfs(s, t, INF, G);
    if(f == 0) break;
    ans += f;
  }
  cout << ans << endl;
}

