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;
int n, m;
int head[5010], deg[5010];
struct Edge {
  int from;
  int to;
  int next;
} edge[10010];
int cnt;
void add_edge(int from, int to) {
  deg[to]++;
  cnt++;
  edge[cnt].from = from;
  edge[cnt].to = to;
  edge[cnt].next = head[from];
  head[from] = cnt;
}
int ans;
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    add_edge(a, b);
  }
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (deg[i] == 0) q.push(i);
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    ans++;
    for (int e = head[t]; e; e = edge[e].next) {
      int to = edge[e].to;
      deg[to]--;
      if (deg[to] == 0) q.push(to);
    }
  }
  if (ans != n) {
    cout << 2 << endl;
    for (int i = 1; i <= m; i++) cout << (edge[i].from > edge[i].to) + 1 << " ";
  } else {
    cout << 1 << endl;
    for (int i = 1; i <= m; i++) cout << 1 << " ";
  }
}