Educational Codeforces Round 72 (Rated for Div. 2)D. Coloring Edges 神仙规律

D. Coloring Edges

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a directed graph with nn vertices and mm directed edges without self-loops or multiple edges.

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

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

Input

The first line contains two integers nn and mm (2≤n≤50002≤n≤5000, 1≤m≤50001≤m≤5000) — the number of vertices and edges in the digraph, respectively.

Next mm lines contain description of edges — one per line. Each edge is a pair of integers uu and vv (1≤u,v≤n1≤u,v≤n, u≠vu≠v) — there is directed edge from uu to vv in the graph.

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

Output

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

In the second line print mm integers c1,c2,…,cmc1,c2,…,cm (1≤ci≤k1≤ci≤k), where cici is a color of the ii-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 kk).

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 

 

题意：

一张有向图，每条边染色，要使得每个环都不能全部为同一种颜色，输出最少需要的颜色和每条边的颜色？

分析：

对于有向图来说，一个环一定包含小号到大号的边和大号到小号的边，所以最多就两种颜色，一看代码就什么都明白了。

 

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int MAXN=50100;
LL n,m,vis[MAXN],in[MAXN],a[MAXN],b[MAXN];
vector<LL> G[MAXN];
int top_sort()
{
    int cnt=0;
    queue<int> q;
    for(int i=1; i<=n; i++)
        if(!in[i])
            q.push(i);
    while(!q.empty())
    {
        int u=q.front();
        cnt++;
        q.pop();
        for(int i=0; i<G[u].size(); i++)
            if(--in[G[u][i]]==0)
                q.push(G[u][i]);
    }
    return (n==cnt);
}
signed main()
{
    cin>>n>>m;
    for(int i=1; i<=m; i++)
    {
        cin>>a[i]>>b[i];
        G[a[i]].push_back(b[i]);
        in[b[i]]++;
    }
    if(top_sort())
    {
        printf("1\n");
        for(int i=1; i<=m; i++)
            printf("1 ");
       cout<<endl;
        return 0;
    }
    printf("2\n");
    for(int i=1; i<=m; i++)
    {
        if(a[i]>b[i])
            printf("1 ");
        else
            printf("2 ");
    }
    cout<<endl;
    return 0;
}