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>
#pragma GCC optimize(2)
using namespace std;
const int inf = (1 << 29);
int n, m;
vector<int> E[5005], R[5005];
int vis[5005];
pair<int, int> e[5005];
bool f;
void calc(int x) {
  if (f) {
    return;
  }
  vis[x] = 1;
  for (int i = 0; i < E[x].size(); ++i) {
    int y = E[x][i];
    if (vis[y] == 1) {
      f = 1;
    } else if (vis[y] == 0) {
      calc(y);
    }
  }
  vis[x] = -1;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    E[x].push_back(y);
    R[y].push_back(x);
    e[i].first = x, e[i].second = y;
  }
  for (int i = 1; i <= n; ++i) {
    if (vis[i] == 0) {
      calc(i);
    }
  }
  if (f == 1) {
    cout << "2\n";
    for (int i = 0; i < m; ++i) {
      if (e[i].first > e[i].second) {
        cout << "2 ";
      } else {
        cout << "1 ";
      }
    }
  } else {
    cout << "1\n";
    for (int i = 0; i < m; ++i) {
      cout << 1 << ' ';
    }
  }
  return 0;
}
