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>
#pragma GCC optimize(3)
using namespace std;
const double PI = acos(-1.0);
const double eps = 1e-6;
struct Edge {
  int to, nxt;
} edge[100005];
int n, m, cnt, head[100005], in[100005], out[100005];
queue<int> q;
pair<int, int> edges[100005];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
void addedge(int u, int v) {
  edge[cnt].to = v;
  edge[cnt].nxt = head[u];
  head[u] = cnt++;
  return;
}
template <typename T>
void debug(T x) {
  cout << "value=" << x << endl;
  return;
}
signed main() {
  memset(head, -1, sizeof(head));
  n = read(), m = read();
  for (int i = 1; i <= m; i++) {
    int u = read(), v = read();
    edges[i].first = u;
    edges[i].second = v;
    addedge(u, v);
    out[u]++;
    in[v]++;
  }
  int tot = 0;
  for (int i = 1; i <= n; i++) {
    if (in[i] == 0) {
      q.push(i);
      tot++;
    }
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int k = head[u]; k != -1; k = edge[k].nxt) {
      int v = edge[k].to;
      in[v]--;
      if (!in[v]) {
        q.push(v);
        tot++;
      }
    }
  }
  if (tot == n) {
    cout << 1 << endl;
    for (int i = 1; i <= m; i++) printf("1 ");
  } else {
    cout << 2 << endl;
    for (int i = 1; i <= m; i++) {
      if (edges[i].first < edges[i].second)
        printf("1 ");
      else
        printf("2 ");
    }
  }
  return 0;
}