D. Coloring Edges

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≤≤50002≤n≤5000, 1≤≤50001≤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≤,≤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 1,2,…,c1,c2,…,cm (1≤≤1≤ci≤k), where ci 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

Copy

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

output

Copy

1
1 1 1 1 1

input

Copy

3 3
1 2
2 3
3 1

output

Copy

2
1 1 2

 题解:有向图性质:若有环,则环中必有从编号小的点指向编号大的点,也有编号大的点指向编号小的点.

则若没有环,则都染成1即可,否则只需将从小指向大的边染为1,否则染为2即可.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=100010;
vector<int>G[maxn];
int flag;
int u[maxn],v[maxn],vis[maxn];
void DFS(int u)
{
  if(flag)return ;
  vis[u]=1;//正在访问
  for(int i=0;i<G[u].size();i++){
    int v=G[u][i];
    if(vis[v]==0)DFS(v);//没访问过
    else if(vis[v]==1){//下一个节点正在访问,即有环
      flag=1;
      return ;
    }
  }
  vis[u]=2;//访问结束
}
int main()
{
  int n,m;
  cin>>n>>m;
  for(int i=1;i<=m;i++){
    cin>>u[i]>>v[i];
    G[u[i]].push_back(v[i]);
  }
  for(int i=1;i<=n;i++){
    if(!vis[i]){
      DFS(i);
    }
  }
  if(!flag){
    cout<<1<<endl;
    for(int i=1;i<=m;i++)cout<<1<<" ";
    cout<<endl;
  }
  else{
    cout<<2<<endl;
    for(int i=1;i<=m;i++){
      if(u[i]<v[i])cout<<1<<" ";
      else cout<<2<<" ";
    }
    cout<<endl;
  }
  return 0;
}