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 int N = 1e5 + 6;
const long long M = 1e9 + 7;
vector<int> g[N];
int vis[N], flag, now;
void dfs_cycle(int node) {
  vis[node] = 1;
  for (int it : g[node]) {
    if (flag) return;
    if (vis[it] == 1) {
      flag = 1;
      return;
    } else if (vis[it] == 0)
      dfs_cycle(it);
  }
  vis[node] = 2;
}
int main() {
  int n, m;
  scanf("%d", &n);
  scanf("%d", &m);
  int arr[m][2];
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d", &x);
    scanf("%d", &y);
    g[x].push_back(y);
    arr[i][0] = x;
    arr[i][1] = y;
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs_cycle(i);
    }
  }
  if (flag == 0) {
    printf("%d\n", 1);
    for (int i = 0; i < m; i++) printf("1 ");
    return 0;
  }
  printf("%d\n", 2);
  for (int i = 0; i < m; i++) {
    if (arr[i][0] < arr[i][1])
      printf("1 ");
    else
      printf("2 ");
  }
}
