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;
long long n, m, x[5005], y[5005];
long long ee, h[5005], nex[5005], to[5005];
long long col[5005], du[5005];
void addedge(long long x, long long y) {
  nex[++ee] = h[x];
  to[ee] = y;
  h[x] = ee;
}
bool topo() {
  queue<long long> q;
  for (long long i = 1; i <= n; i++)
    if (du[i] == 0) q.push(i);
  long long s = 0;
  while (!q.empty()) {
    long long x = q.front();
    q.pop();
    s++;
    for (long long i = h[x]; i; i = nex[i])
      if ((--du[to[i]]) == 0) q.push(to[i]);
  }
  return s == n;
}
signed main() {
  cin >> n >> m;
  for (long long i = 1; i <= m && cin >> x[i] >> y[i]; i++)
    addedge(x[i], y[i]), du[y[i]]++;
  if (topo()) {
    cout << 1 << endl;
    for (long long i = 1; i <= m; i++) cout << 1 << ' ';
  } else {
    cout << 2 << endl;
    for (long long i = 1; i <= m; i++)
      if (x[i] > y[i])
        cout << 1 << ' ';
      else
        cout << 2 << ' ';
  }
  return puts(""), 0;
}
