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>
const double PI = acos(-1.0);
const int maxn = 1e6;
const int INF = 0x3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
using namespace std;
struct node {
  int a, b, color;
} edge[maxn];
int n, m, ans;
int vis[maxn];
vector<int> v[maxn];
void dfs(int x, int fa) {
  if (vis[x]) {
    if (x == fa) ans = 2;
    return;
  }
  vis[x] = 1;
  for (int i = 0; i < v[x].size(); i++) dfs(v[x][i], fa);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    edge[i].a = a;
    edge[i].b = b;
    edge[i].color = 1;
    v[a].push_back(b);
  }
  ans = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) vis[j] = 0;
    dfs(i, i);
  }
  cout << ans << endl;
  if (ans == 1) {
    for (int i = 0; i < m; i++) cout << 1 << " ";
    cout << endl;
  } else {
    for (int i = 1; i <= m; i++) {
      if (edge[i].a < edge[i].b) edge[i].color = 2;
    }
    for (int i = 1; i <= m; i++) cout << edge[i].color << " ";
    cout << endl;
  }
  return 0;
}