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;
int vis[5555], dp[5555], vis2[5555], tt;
vector<pair<int, int> > v[5555];
void recur(int now) {
  vis[now] = 1;
  vis2[now] = 1;
  for (int e = 0; e < v[now].size(); e++) {
    int next = v[now][e].first;
    int nextn = v[now][e].second;
    if (vis[next] == 1) {
      dp[nextn] = 2;
      tt = 1;
    } else {
      dp[nextn] = 1;
      if (vis2[next] == 0) recur(next);
    }
  }
  vis[now] = 0;
}
int main(void) {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int e = 0; e < m; e++) {
    int a, b;
    scanf("%d%d", &a, &b);
    v[a].push_back(make_pair(b, e));
  }
  for (int e = 1; e <= n; e++) {
    if (vis2[e] == 0) {
      recur(e);
    }
  }
  if (tt == 1) {
    printf("2\n");
    for (int e = 0; e < m; e++) printf("%d ", dp[e]);
  } else {
    printf("1\n");
    for (int e = 0; e < m; e++) printf("%d ", dp[e]);
  }
}