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 
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int dx[8] = {0, -1, 0, 1, -1, -1, 1, 1},
          dy[8] = {1, 0, -1, 0, -1, 1, -1, 1};
long long qk(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x;
    y >>= 1;
    x = x * x;
  }
  return ans;
}
vector<int> vt[5005];
int vis[5005];
int k = 0;
int xx[5005], yy[5005];
int cnt;
void dfs(int x) {
  if (vis[x]) {
    if (x == cnt) k = 2;
    return;
  }
  vis[x] = 1;
  for (auto it : vt[x]) {
    dfs(it);
  }
}
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> xx[i] >> yy[i];
    vt[xx[i]].push_back(yy[i]);
  }
  k = 1;
  for (int i = 1; i <= n; i++) {
    memset(vis, 0, sizeof vis);
    cnt = i;
    dfs(i);
  }
  if (k == 1) {
    cout << 1 << endl;
    for (int i = 1; i <= m; i++) {
      cout << 1 << ' ';
    }
  } else {
    cout << 2 << endl;
    for (int i = 1; i <= m; i++) {
      if (xx[i] > yy[i])
        cout << 1 << ' ';
      else {
        cout << 2 << ' ';
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  ;
  solve();
  return 0;
}
