Problem: You are given a directed graph with n vertices and m directed edges without self-loops or multiple edges.

Let's denote the k-coloring of a digraph as following: you color each edge in one of k colors. The k-coloring is good if and only if there no cycle formed by edges of same color.

Find a good k-coloring of given digraph with minimum possible k.

Input

The first line contains two integers n and m (2 ≤ n ≤ 5000, 1 ≤ m ≤ 5000) — the number of vertices and edges in the digraph, respectively.

Next m lines contain description of edges — one per line. Each edge is a pair of integers u and v (1 ≤ u, v ≤ n, u ≠ v) — there is directed edge from u to v in the graph.

It is guaranteed that each ordered pair (u, v) appears in the list of edges at most once.

Output

In the first line print single integer k — the number of used colors in a good k-coloring of given graph.

In the second line print m integers c_1, c_2, ..., c_m (1 ≤ c_i ≤ k), where c_i is a color of the i-th edge (in order as they are given in the input).

If there are multiple answers print any of them (you still have to minimize k).

Examples

Input


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


Output


1
1 1 1 1 1 


Input


3 3
1 2
2 3
3 1


Output


2
1 1 2 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (a < b) return gcd(b, a);
  return b == 0 ? a : gcd(b, a % b);
}
const int N = 5e3 + 9;
int in[N], h[N], ans[N];
int cnt = 0, n, m;
bool vis[N];
bool vis1[N];
struct edge {
  int to, nex, pre;
} e[N];
struct node {
  int u, col;
} a[N];
void add(int u, int v) {
  e[++cnt] = (edge){v, h[u], u};
  h[u] = cnt;
}
bool judge() {
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      q;
  for (int i = 1; i <= n; ++i) q.push(make_pair(in[i], i));
  bool ok = 1;
  while (!q.empty()) {
    int num = q.top().first;
    int u = q.top().second;
    q.pop();
    if (vis1[u]) continue;
    vis1[u] = 1;
    if (num) {
      ok = 0;
      break;
    }
    for (int i = h[u]; i; i = e[i].nex) {
      int v = e[i].to;
      if (vis1[v]) continue;
      --in[v];
      q.push(make_pair(in[v], v));
    }
  }
  return ok;
}
void dfs(int x) {
  int u = e[x].to;
  for (int i = h[u]; i; i = e[i].nex) {
    int v = e[i].to;
    if (vis[i]) continue;
    vis[i] = 1;
    if (v > u)
      ans[i] = 1;
    else
      ans[i] = 2;
    dfs(i);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    ++in[v];
    add(u, v);
  }
  if (judge()) {
    puts("1");
    for (int i = 1; i <= m; ++i) printf("%d ", 1);
    return 0;
  }
  for (int i = 1; i <= m; ++i) {
    if (!vis[i]) {
      vis[i] = 1;
      if (e[i].to > e[i].pre)
        ans[i] = 1;
      else
        ans[i] = 2;
      dfs(i);
    }
  }
  puts("2");
  for (int i = 1; i <= m; ++i) printf("%d ", ans[i]);
  return 0;
}