Problem: You are given an undirected bipartite graph without multiple edges. You should paint the edges of graph to minimal number of colours, so that no two adjacent edges have the same colour.

Input

The first line contains three integers a, b, m (1 ≤ a, b ≤ 1000, 0 ≤ m ≤ 105), a is the size of the first part, b is the size of the second part, m is the number of edges in the graph.

Each of the next m lines contains two integers x, y (1 ≤ x ≤ a, 1 ≤ y ≤ b), where x is the number of the vertex in the first part and y is the number of the vertex in the second part. It is guaranteed that there are no multiple edges.

Output

In the first line print integer c — the minimal number of colours. The second line should contain m integers from 1 to c — the colours of the edges (in the order they appear in the input).

If there are several solutions, you can print any one of them.

Examples

Input

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


Output

3
1 2 3 1 2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(0.0) * 2.0;
const long double eps = 1e-10;
const int step[8][2] = {{-1, 0}, {0, 1}, {1, 0},  {0, -1},
                        {-1, 1}, {1, 1}, {1, -1}, {-1, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <typename t, typename t1>
t min1(t a, t1 b) {
  return a < b ? a : b;
}
template <typename t, typename... arg>
t min1(t a, arg... arr) {
  return min1(a, min1(arr...));
}
template <typename t, typename t1>
t max1(t a, t1 b) {
  return a > b ? a : b;
}
template <typename t, typename... arg>
t max1(t a, arg... arr) {
  return max1(a, max1(arr...));
}
inline int jud(double a, double b) {
  if (abs(a) < eps && abs(b) < eps)
    return 0;
  else if (abs1(a - b) / max(abs1(a), abs1(b)) < eps)
    return 0;
  if (a < b) return -1;
  return 1;
}
template <typename t>
inline int jud(t a, t b) {
  if (a < b) return -1;
  if (a == b) return 0;
  return 1;
}
template <typename it, typename t1>
inline int find(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1) {
  if (na == 0) return 0;
  int be = 0, en = na - 1;
  if (*a <= *(a + na - 1)) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(*(a + be), val) == 1) be--;
    if (!f_small && jud(*(a + be), val) == -1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(*(a + be), val) == -1) be--;
    if (f_small && jud(*(a + be), val) == 1) be++;
  }
  return be;
}
template <class T>
inline T lowb(T num) {
  return num & (-num);
}
inline int bitnum(unsigned int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(unsigned long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitnum(long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitmaxl(unsigned int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(unsigned long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
inline int bitmaxl(long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
long long pow(long long n, long long m, long long mod = 0) {
  if (m < 0) return 0;
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
template <class t1, class t2>
inline void add(t1 &a, t2 b, int mod = -1) {
  if (mod == -1) mod = 1000000007;
  a += b;
  while (a >= mod) a -= mod;
  while (a < 0) a += mod;
}
template <class t>
void output1(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) cerr << arr[i] << ' ';
  cerr << endl;
}
template <class t>
void output2(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) output1(arr[i]);
}
const int maxn = 4010;
const int maxm = 200100;
struct edge {
  int to, nxt, pre;
} e[maxm * 2];
int le, head[maxn];
int match[maxn], me[maxn];
int st[maxn], in[maxn];
int a, b, m;
int ans[maxm];
int du[maxn], du1[maxn], rot, rans;
void addedge(int a, int b) {
  e[le].to = b;
  e[le].nxt = head[a];
  if (head[a] != -1) e[head[a]].pre = le;
  e[le].pre = -1;
  head[a] = le++;
}
inline void era(int p) {
  int a = e[p].to, b = e[p ^ 1].to;
  if (e[p].pre != -1)
    e[e[p].pre].nxt = e[p].nxt;
  else
    head[b] = e[head[b]].nxt;
  if (e[p].nxt != -1) e[e[p].nxt].pre = e[p].pre;
  p ^= 1;
  if (e[p].pre != -1)
    e[e[p].pre].nxt = e[p].nxt;
  else
    head[a] = e[head[a]].nxt;
  if (e[p].nxt != -1) e[e[p].nxt].pre = e[p].pre;
}
int dfs(int no) {
  if (in[no]) return 0;
  in[no] = 1;
  for (int i = st[no]; i != -1; i = e[i].nxt) {
    if (match[e[i].to] == no) continue;
    if (du[match[e[i].to]] < rans - rot + 1) {
      match[match[e[i].to]] = -1;
      match[e[i].to] = no;
      match[no] = e[i].to;
      me[no] = me[e[i].to] = i;
      return 1;
    }
    if (match[e[i].to] == -1 || dfs(match[e[i].to])) {
      match[e[i].to] = no;
      match[no] = e[i].to;
      me[no] = me[e[i].to] = i;
      return 1;
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  memset(head, -1, sizeof(head));
  scanf("%d%d%d", &a, &b, &m);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    addedge(u, v + a);
    addedge(v + a, u);
    du[u]++;
    du[v + a]++;
  }
  rans = 0;
  for (int i = 0; i < a + b; i++) rans = max(rans, du[i]);
  for (int i = 1; i <= rans; i++) {
    rot = i;
    memset(match, -1, sizeof(match));
    for (int j = 0; j < a + b; j++) st[j] = head[j];
    for (int j = 0; j < a + b; j++)
      if (du[j] == rans - i + 1)
        if (match[j] == -1) {
          memset(in, 0, sizeof(in));
          int flag = dfs(j);
          if (!flag) {
            cerr << i << endl;
          }
        }
    for (int j = 0; j < a + b; j++)
      if (match[j] > j) {
        du[match[j]]--;
        du[j]--;
        era(me[j]);
        ans[me[j] / 2] = i;
      }
  }
  printf("%d\n", rans);
  for (int i = 0; i < m; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}