#include <bits/stdc++.h>

using namespace std;
const int N = 100010, M = 2000010; // 因为要建新图，两倍的边
int n, m, mod;                     // 点数、边数、取模的数

int f[N], g[N];

int h[N], hs[N], e[M], ne[M], idx; // h: 原图；hs: 新图
void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

// tarjan求强连通分量
int dfn[N], low[N], ts, in_stk[N], stk[N], top;
int id[N], scc_cnt, sz[N];
void tarjan(int u) {
    dfn[u] = low[u] = ++ts;
    stk[++top] = u;
    in_stk[u] = 1;
    for (int i = h[u]; ~i; i = ne[i]) {
        int v = e[i];
        if (!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if (in_stk[v])
            low[u] = min(low[u], low[v]);
    }
    if (dfn[u] == low[u]) {
        ++scc_cnt;
        int x;
        do {
            x = stk[top--];
            in_stk[x] = 0;
            id[x] = scc_cnt;
            sz[scc_cnt]++;
        } while (x != u);
    }
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("1175_Preapre.in", "r", stdin);
#endif
    memset(h, -1, sizeof h); // 原图

    scanf("%d %d", &n, &m);

    while (m--) {
        int a, b;
        scanf("%d %d", &a, &b);
        add(a, b);
    }
    // (1) tarjan算法求强连通分量
    for (int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i);

    for (int i = 1; i <= n; i++)
        cout << "id[" << i << "]=" << id[i] << " ";
    cout << endl;

    cout << "scc_cnt=" << scc_cnt << endl;

    return 0;
}