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;
const int inf = 0x3f3f3f3f;
const int maxn = 5e3 + 10;
vector<int> topo;
int n, m;
vector<int> G[maxn];
int mp[maxn][maxn];
int x[maxn], y[maxn];
int f = 1;
bool topoSort() {
  int indgree[maxn] = {0};
  int visit[maxn] = {0};
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < G[i].size(); j++) {
      indgree[G[i][j]]++;
    }
  }
  int control = 0;
  while (control < n) {
    int u = 0, i;
    for (i = 0; i < n; i++) {
      if (!visit[i] && indgree[i] == 0) {
        u = i;
        break;
      }
    }
    if (i == n) {
      return false;
    }
    visit[u] = 1;
    topo.push_back(u);
    for (int j = 0; j < G[u].size(); j++) {
      indgree[G[u][j]]--;
    }
    control++;
  }
  return true;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      mp[i][j] = 0;
      mp[j][i] = 1;
    }
  }
  for (int i = 0; i < m; i++) {
    cin >> x[i] >> y[i];
    x[i]--;
    y[i]--;
    G[x[i]].push_back(y[i]);
  }
  if (topoSort()) {
    cout << 1 << endl;
    for (int i = 0; i < m; i++) {
      cout << 1 << ' ';
    }
  } else {
    cout << 2 << endl;
    for (int i = 0; i < m; i++) {
      cout << mp[x[i]][y[i]] + 1 << ' ';
    }
  }
}