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;
struct Edge {
  int to, nxt;
} e[5005];
queue<int> q;
int n, m, h[5005], num, d[5005];
bool vis[5005], sel[5005];
void add(int u, int v) {
  e[++num].to = v;
  e[num].nxt = h[u];
  h[u] = num;
  d[v]++;
}
bool topo() {
  int i, u, v;
  for (i = 1; i <= n; i++)
    if (!d[i]) q.push(i);
  while (!q.empty()) {
    u = q.front();
    q.pop();
    for (i = h[u]; i; i = e[i].nxt) {
      v = e[i].to;
      if (!--d[v]) q.push(v);
    }
  }
  for (i = 1; i <= n; i++)
    if (d[i]) return 0;
  return 1;
}
void dfs(int u) {
  vis[u] = 1;
  int v, i;
  for (i = h[u]; i; i = e[i].nxt) {
    v = e[i].to;
    if (!vis[v]) {
      sel[i] = 1;
      dfs(v);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  int i, u, v;
  cin >> n >> m;
  for (i = 1; i <= m; i++) {
    cin >> u >> v;
    add(u, v);
    if (u > v) sel[i] = 1;
  }
  if (topo()) {
    cout << 1 << endl;
    for (i = 1; i <= m; i++) cout << "1 ";
    return 0;
  }
  cout << 2 << endl;
  for (i = 1; i <= m; i++) cout << 2 - sel[i] << " ";
  return 0;
}