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 = 5010;
int n, m;
int tot, bian[N], nxt[N], head[N], rd[N];
void add(int x, int y) {
  ++tot, bian[tot] = y, nxt[tot] = head[x], head[x] = tot, rd[y]++;
}
int ans = 1;
int c[N];
int v[N];
void dfs(int x) {
  v[x] = 1;
  for (int i = head[x]; i; i = nxt[i]) {
    int y = bian[i];
    if (v[y]) {
      if (v[y] == 1) {
        c[i] = 2;
        ans = 2;
      } else {
        c[i] = 1;
      }
    } else {
      c[i] = 1;
      dfs(y);
    }
  }
  v[x] = 2;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
  }
  for (int i = 1; i <= n; ++i) {
    if (v[i]) continue;
    dfs(i);
  }
  cout << ans << endl;
  for (int i = 1; i <= m; ++i) {
    printf("%d ", c[i]);
  }
}
