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(2)
using namespace std;
const long long inf = 1e18 + 7;
typedef struct {
  int u, v;
} po;
int a, b, n, k, c, book, m, P, x, y, t, cnt, book1, book2;
int dit[5][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}, {0, 0}};
int ar[5010][5010];
int indeg[5010];
po edg[5010];
queue<int> q;
void topo_sort() {
  for (int i = 1; i <= n; i++)
    if (indeg[i] == 0) q.push(i);
  cnt = 0;
  while (!q.empty()) {
    const int u = q.front();
    q.pop();
    cnt++;
    for (int i = 1; i <= n; i++)
      if (ar[u][i]) {
        indeg[i]--;
        if (indeg[i] == 0) q.push(i);
      }
  }
  if (cnt == n) {
    cout << 1 << endl;
    for (int i = 1; i <= m; i++) cout << 1 << " ";
    cout << endl;
  } else {
    cout << 2 << endl;
    for (int i = 1; i <= m; i++) {
      if (edg[i].u > edg[i].v)
        cout << 2 << " ";
      else
        cout << 1 << " ";
    }
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  while (cin >> n >> m) {
    memset(indeg, 0, sizeof(indeg));
    po tt;
    for (int i = 1; i <= m; i++) {
      cin >> tt.u >> tt.v;
      edg[i] = tt;
      ar[tt.u][tt.v] = 1;
      indeg[tt.v]++;
    }
    topo_sort();
  }
  return 0;
}