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 
#include <bits/stdc++.h>
using namespace std;
const int N = 5005;
int vis[N];
vector<pair<int, int> > adj[N];
int color[N];
void dfs(int node) {
  vis[node] = 1;
  int sz = adj[node].size();
  for (int i = 0; i < sz; ++i) {
    int eid = adj[node][i].first;
    int ch = adj[node][i].second;
    if (vis[ch] == 0) {
      dfs(ch);
    } else if (vis[ch] == 1) {
      color[eid] = 1;
    }
  }
  vis[node] = 2;
}
void solve() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    adj[a].push_back(make_pair(i, b));
  }
  for (int i = 1; i <= n; ++i) {
    if (vis[i] == 0) {
      dfs(i);
    }
  }
  int maxColor = 0;
  for (int i = 0; i < m; ++i) {
    maxColor = max(maxColor, color[i] + 1);
  }
  printf("%d\n", maxColor);
  for (int i = 0; i < m; ++i) {
    printf("%d ", color[i] + 1);
  }
}
int main() { solve(); }
