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>
const int N = 5e3 + 10;
int head[N], edge[N], ver[N], nex[N];
int tot = 0;
void addEdge(int x, int y) {
  ver[++tot] = y;
  nex[tot] = head[x];
  head[x] = tot;
}
int n, m, xx[N], yy[N];
bool flag = false;
int vis[N];
void dfs(int x) {
  if (flag) return;
  vis[x] = 1;
  for (int i = head[x]; i; i = nex[i]) {
    int y = ver[i];
    if (vis[y] == 1) {
      flag = true;
      return;
    } else if (!vis[y])
      dfs(y);
  }
  vis[x] = 2;
}
void solve() {
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  if (flag) {
    puts("2");
    for (int i = 1; i <= m; i++) {
      if (xx[i] > yy[i])
        printf("2 ");
      else
        printf("1 ");
    }
  } else {
    puts("1");
    for (int i = 1; i <= m; i++) printf("1 ");
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &xx[i], &yy[i]);
    addEdge(xx[i], yy[i]);
  }
  solve();
  return 0;
}
