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 n, m, ans, cnt;
const int N = 3e5 + 3;
int a[N], du[N], tp[N], ehead[N];
char s[N];
struct ed {
  int u, to, nxt;
} e[N];
vector<int> ee[N];
queue<int> q;
inline void add(int u, int v) {
  e[++cnt].to = v;
  e[cnt].u = u;
  e[cnt].nxt = ehead[u];
  ehead[u] = cnt;
}
int main() {
  cnt = 0;
  cin >> n >> m;
  int u, v;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v;
    add(u, v);
    du[v]++;
  }
  for (int i = 1; i <= n; i++) {
    if (!du[i]) q.push(i);
  }
  while (!q.empty()) {
    u = q.front();
    q.pop();
    for (int i = ehead[u]; i; i = e[i].nxt) {
      v = e[i].to;
      du[v]--;
      if (du[v] == 0) q.push(v);
    }
  }
  bool flag = true;
  for (int i = 1; i <= n; i++)
    if (du[i] != 0) flag = false;
  if (flag) {
    ans = 1;
  }
  if (ans != 1) ans = 2;
  if (ans == 1) {
    puts("1");
    for (int i = 1; i <= m; i++) printf("1 ");
  } else {
    cout << 2 << endl;
    for (int i = 1; i <= m; i++) {
      if (e[i].u < e[i].to)
        cout << "1 ";
      else
        cout << "2 ";
    }
  }
  return 0;
}