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;
vector<int> g[5005];
int u[5005];
int v[5005];
int vis[5005];
bool flag = false;
void dfs(int u) {
  if (flag) return;
  vis[u] = 1;
  for (int i = 0; i < (int)g[u].size(); i++) {
    int v = g[u][i];
    if (!vis[v])
      dfs(v);
    else if (vis[v] == 1) {
      flag = true;
      return;
    }
  }
  vis[u] = 2;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &u[i], &v[i]);
    g[u[i]].push_back(v[i]);
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(u[i]);
    }
    if (flag) break;
  }
  if (!flag) {
    cout << 1 << endl;
    cout << 1;
    for (int i = 2; 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;
  }
}
