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 len = 2e5 + 10;
const double pi = acos(-1.0);
const long long mod = 998244353;
const int inf = 0x3f3f3f3f;
const int maxn = 1e9 + 5;
int sum[len];
int vis[5005], f;
vector<int> g[len];
struct node {
  int x, y;
} s[len];
void dfs(int x) {
  vis[x] = 0;
  for (int i = 0; i < g[x].size(); i++) {
    int v = g[x][i];
    if (vis[v] == 0) {
      f = 1;
      return;
    } else if (vis[v] == -1)
      dfs(v);
  }
  vis[x] = 1;
  return;
}
int main() {
  int n, m;
  memset(vis, -1, sizeof(vis));
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    s[i].x = x;
    s[i].y = y;
    sum[y]++;
  }
  for (int i = 1; i <= m; i++)
    if (sum[s[i].x] == 0) {
      memset(vis, -1, sizeof(vis));
      dfs(s[i].x);
    }
  dfs(s[1].x);
  if (f) {
    cout << 2 << endl;
    for (int i = 1; i <= m; i++)
      if (s[i].x > s[i].y)
        cout << 1 << " ";
      else
        cout << 2 << " ";
  } else {
    cout << 1 << endl;
    for (int i = 1; i <= m; i++) cout << 1 << " ";
  }
}