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 = 1e4 + 10;
const long long MOD = 1e9 + 3;
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline int lb(int x) { return x & -x; }
long long qpow(long long base, int n) {
  if (base == 0) return 0;
  long long ans = 1;
  while (n) {
    if (n & 1) ans *= base;
    n >>= 1;
    base *= base;
  }
  return ans;
}
int n, m, k = 1;
stack<int> st;
struct Edge {
  int from, to, next;
} e[maxn];
int cnt, head[maxn], in[maxn], col[maxn];
inline void addedge(int u, int v) {
  e[++cnt].to = v;
  e[cnt].next = head[u];
  e[cnt].from = u;
  head[u] = cnt;
  ++in[v];
}
int cnt_point;
void topo() {
  queue<int> qu;
  for (int i = (1); i <= (n); ++i)
    if (!in[i]) qu.push(i);
  while (!qu.empty()) {
    int tmp = qu.front();
    qu.pop();
    ++cnt_point;
    for (int i = head[tmp]; ~i; i = e[i].next) {
      --in[e[i].to];
      if (!in[e[i].to]) qu.push(e[i].to);
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  memset(head, -1, sizeof(head));
  for (int i = 0, x, y; i < m; ++i) {
    scanf("%d%d", &x, &y);
    addedge(x, y);
  }
  topo();
  if (cnt_point == n) {
    for (int i = (1); i <= (m); ++i) col[i] = 1;
  } else {
    k = 2;
    for (int i = (1); i <= (m); ++i) col[i] = e[i].from < e[i].to ? 1 : 2;
  }
  printf("%d\n", k);
  for (int i = (1); i <= (m); ++i) printf("%d%c", col[i], i == m ? '\n' : ' ');
}