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;
long long input() {
  long long x = 0, f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= ch == '-', ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
const int N = 5007;
struct edge {
  int v, next, id;
} e[N];
int head[N], cnt;
int d[N];
int n, m, k = 1, f = 0;
int Ans[N];
int vis[N];
void Ins(int u, int v, int id) {
  e[++cnt] = (edge){v, head[u], id}, head[u] = cnt;
  d[v]++;
}
void topo() {
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (!vis[i] && d[i] == 0) q.push(i), vis[i] = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = head[u]; i; i = e[i].next) {
      Ans[e[i].id] = 1;
      if (vis[e[i].v]) continue;
      d[e[i].v]--;
      if (!d[e[i].v]) {
        q.push(e[i].v);
        vis[e[i].v] = 1;
      }
    }
  }
}
void check() {
  topo();
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      for (int j = head[i]; j; j = e[j].next) {
        k = 2;
        Ans[e[j].id] = e[j].v > i ? 1 : 2;
        f = 0;
        d[e[j].v]--;
      }
      vis[i] = 1;
    }
  }
}
int main() {
  n = input(), m = input();
  for (int i = 1; i <= m; i++) {
    int u = input(), v = input();
    Ins(u, v, i);
  }
  check();
  printf("%d\n", k);
  for (int i = 1; i <= m; i++) {
    printf("%d%c", Ans[i], i == m ? '\n' : ' ');
  }
}