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;
const int maxn = (int)1e5 + 7;
struct Edge {
  int from, nex, to;
} edge[maxn << 1];
int head[maxn], cnt;
inline void addedge(int u, int v) {
  edge[cnt] = {u, head[u], v};
  head[u] = cnt++;
}
int dfn[maxn], low[maxn], st[maxn], topp, indexx, scc[maxn], sccnum;
void tarjan(int root) {
  dfn[root] = low[root] = ++indexx;
  st[++topp] = root;
  for (int i = head[root]; ~i; i = edge[i].nex) {
    int v = edge[i].to;
    if (!dfn[v]) {
      tarjan(v);
      low[root] = min(low[root], low[v]);
    } else if (!scc[v])
      low[root] = min(low[root], dfn[v]);
  }
  if (low[root] == dfn[root]) {
    sccnum++;
    for (;;) {
      int x = st[topp--];
      scc[x] = sccnum;
      if (x == root) break;
    }
  }
}
int ans[maxn];
int main() {
  int n, m;
  memset(head, -1, sizeof head);
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    addedge(u, v);
  }
  for (int i = 1; i <= n; i++) {
    if (!dfn[i]) {
      tarjan(i);
    }
  }
  if (sccnum == n) {
    cout << 1 << endl;
    for (int i = 1; i <= m; i++) {
      cout << 1;
      if (i == m)
        cout << endl;
      else
        cout << " ";
    }
    return 0;
  }
  cout << 2 << endl;
  for (int i = 0; i < cnt; i++) {
    if (edge[i].from < edge[i].to) {
      ans[i] = 1;
    } else
      ans[i] = 2;
  }
  for (int i = 0; i < cnt; i++) {
    cout << ans[i];
    if (i == cnt - 1)
      cout << endl;
    else
      cout << " ";
  }
  return 0;
}