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 PYTHON3):

Solution:
from collections import deque

class HopcroftKarp:
    def __init__(self, N0, N1):
        self.N0 = N0; self.N1 = N1
        self.N = N = 2 + N0 + N1
        self.E = E = [[] for _ in range(N)]
        for end in range(N0): # vertex 0 is the source
            v = end + 2
            E[0].append([v, 1, len(E[v])])
            E[v].append([0, 0, len(E[0]) - 1])
        for init in range(N1): # vertex 1 is the sink
            u = init + N0 + 2
            E[u].append([1, 1, len(E[1])])
            E[1].append([u, 0, len(E[u]) - 1])
    
    def add_edge(self, p1, p2):
        '''
            e[0] = the head vertex of e,
            e[1] = current capacity of e,
            e[2] = the index of reverse edge of e.
        '''
        p1 += 2; p2 += self.N0 + 2
        self.E[p1].append([p2, 1, len(self.E[p2])])
        self.E[p2].append([p1, 0, len(self.E[p1]) - 1])
    
    def _exist_augpath(self):
        self.level = level = [-1] * self.N # the level (depth) of each vertex
        E = self.E
        q = deque([(0, 0)]) # (vertex, level)
        while q:
            v, lv = q.popleft()
            if level[v] < 0: # visited v for the first time
                level[v] = lv
                for u, cap, _ in E[v]:
                    if cap == 0 or level[u] >= 0: continue
                    q.append((u, lv + 1))
        return level[1] >= 0
    
    def _blocking_flow(self):
        s = 0; t = 1
        aug_flow = 0
        finished = [False] * self.N
        parent = [-1] * self.N
        revs = [-1] * self.N
        bottleneck = -1; temp_flow = 0
        level, E = self.level, self.E
        stack = [(s, -1, float('inf'), -1, 0)] # (vertex, parent, flow, rev_idx, status)
        while stack:
            v, p, f, rev, st = stack.pop()
            if temp_flow and v != bottleneck: continue
            if st == 0 and not finished[v]:
                parent[v] = p; revs[v] = rev
                if temp_flow: # offset
                    f -= temp_flow; temp_flow = 0; bottleneck = -1
                if v == t: # an augment path is found
                    aug_flow += f
                    if p != -1: stack += [(v, p, f, rev, 3)]
                    continue
                n_children = 0
                for u, cap, rev_idx in E[v]:
                    if cap == 0 or finished[u] or level[u] != level[v] + 1: continue
                    if n_children == 0:
                        stack += [(v, p, 0, rev, 2), (u, v, min(f, cap), rev_idx, 0)]
                        n_children += 1
                    else:
                        stack += [(v, p, 0, rev, 1), (u, v, min(f, cap), rev_idx, 0)]
                        n_children += 1
                if n_children == 0: # v is a leaf
                    finished[v] = True
            elif st == 1: # now searching
                continue
            elif st == 2: # search finished
                finished[v] = True
            else: # edge update
                rev_e = E[v][rev]; e = E[p][rev_e[2]]
                e[1] -= f; rev_e[1] += f
                if e[1] == 0: bottleneck = p
                if parent[p] != -1: stack += [(p, parent[p], f, revs[p], 3)]
                else: temp_flow = f
        return aug_flow
    
    def max_matching(self):
        matching = 0
        exist_augpath, blocking_flow = self._exist_augpath, self._blocking_flow
        while True:
            if not exist_augpath(): break
            matching += blocking_flow()
        return matching

N0, N1, M = map(int, input().split())
hk = HopcroftKarp(N0, N1)
for _ in range(M):
    u, v = map(int, input().split())
    hk.add_edge(u, v)
print(hk.max_matching())