Problem: 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

Solution (in CPP):

Solution:

//be naame khodaa
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
#define F first
#define S second
#define pb push_back

inline int in(){int x, y; y = scanf("%d", &x); return x;}

const int N = 404;

vector<int> g[N];
int match[N];
bool mark[N];

bool dfs (int v){
	mark[v] = true;
	for (int u : g[v]){
		if (match[u] == -1 || (!mark[match[u]] && dfs(match[u]))){
			match[v] = u;
			match[u] = v;
			return true;
		} 
	}
	return false;
} 
int maxMatching(int n){
	int ret=0;
	fill(match, match + N, -1);
	while(true)
	{
		fill(mark, mark + n, false);
		int lret = ret;
		for(int i = 0; i < n; i++)
			if(!mark[i] && match[i] == -1)
				ret += dfs(i);
		if(ret == lret)
			break;
	}
	return ret;
}

int main(){
	int n = in(), m = in(), es = in();
	while(es--)
	{
		int u = in(), v = in();
		g[u].push_back(n + v);
	}
	cout << maxMatching(n) << endl;
	return 0;
}