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>
using namespace std;
const long long Inf = 1e18;
const int inf = 0x3f3f3f3f;
const int maxn = 5005;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f * x;
}
struct edgs {
  int from, to, nxt;
} edg[maxn];
int cnt, tot = 1;
int head[maxn];
int vis[maxn];
int ans[maxn];
void add_edg(int from, int to) {
  edg[++cnt].from = from;
  edg[cnt].to = to;
  edg[cnt].nxt = head[from];
  head[from] = cnt;
}
void dfs(int u) {
  vis[u] = 1;
  for (int i = head[u]; i; i = edg[i].nxt) {
    int v = edg[i].to;
    ans[i] = 1;
    if (vis[v] == 1) ans[i] = tot = 2;
    if (!vis[v]) dfs(v);
  }
  vis[u] = 2;
}
int main(void) {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    add_edg(x, y);
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i);
  }
  cout << tot << endl;
  for (int i = 1; i <= m; i++) cout << ans[i] << " ";
  return 0;
}
