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;
const int maxn = 5010;
inline int max(int x, int y) { return x > y ? x : y; }
inline int min(int x, int y) { return x < y ? x : y; }
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 << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return f * x;
}
struct edge {
  int to, pre;
} G[maxn];
int n, m, Len, ma;
int head[maxn], a[maxn];
bool vis[maxn], used[maxn];
void addedge(int x, int y) {
  G[++Len].to = y;
  G[Len].pre = head[x];
  head[x] = Len;
}
void dfs(int u) {
  vis[u] = used[u] = 1;
  int v;
  for (int i = head[u]; i; i = G[i].pre) {
    v = G[i].to;
    if (vis[v]) {
      ma = a[i] = 1;
    } else {
      if (!used[v]) dfs(v);
    }
  }
  vis[u] = 0;
}
int main() {
  ma = Len = 0;
  memset(used, 0, sizeof(used));
  memset(vis, 0, sizeof(vis));
  memset(a, 0, sizeof(a));
  n = read();
  m = read();
  int x, y;
  for (int i = 1; i <= m; ++i) {
    x = read();
    y = read();
    addedge(x, y);
  }
  for (int i = 1; i <= n; ++i)
    if (!used[i]) dfs(i);
  printf("%d\n", ma + 1);
  for (int i = 1; i <= m; ++i) printf("%d ", a[i] + 1);
}