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;
template <class T>
inline bool checkMin(T& a, T b) {
  return (a > b ? a = b, 1 : 0);
}
template <class T>
inline bool checkMax(T& a, T b) {
  return (a < b ? a = b, 1 : 0);
}
const int N = 5005;
vector<int> e[N];
int n, m;
void init() { scanf("%d%d", &n, &m); }
int vis[N];
void dfs(int u) {
  if (vis[u]) return;
  vis[u] = 1;
  for (int i = 0; i < (int)e[u].size(); i++) {
    int vv = e[u][i];
    dfs(vv);
  }
}
int ss[N];
void solve() {
  int ans = 1;
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    memset(vis, 0, sizeof(vis));
    dfs(v);
    if (vis[u]) {
      ans = 2;
      ss[i] = ans;
    } else {
      e[u].push_back(v);
      ss[i] = 1;
    }
  }
  printf("%d\n", ans);
  for (int i = 0; i < m; i++) printf("%d ", ss[i]);
}
int main() {
  init();
  solve();
  return 0;
}
