用状态压缩来表示集合，不妨设 $f(x)$ 表示从 $x$ 出发能够到达的点构成的集合
$$ f(x) = \{x\} \cup \left( \bigcup_{(x, y) \text{存在有向边}} f(y)\right) $$

const int maxn = 30000 + 10;
int head[maxn], ne[maxn], ver[maxn], deg[maxn], n, m, tot = 0;
vector<int> a;

void add(int x, int y) {
    ver[++tot] = y; ne[tot] = head[x]; head[x] = tot;
    deg[y]++;
}

void toposort() {
    queue<int> q;
    for (int i = 1; i <= n; i++) {
        if (deg[i] == 0) q.push(i);
    }

    while (q.size()) {
        int x = q.front(); q.pop();
        a.push_back(x);
        for (int i = head[x]; i; i = ne[i]) {
            int y = ver[i];
            if (--deg[y] == 0) q.push(y);
        }
    }
}

bitset<maxn> f[maxn];
void solve() {
    for (int j = a.size()-1; j >= 0; j--) {
        int x = a[j];
        f[x][x] = 1;
        for (int i = head[x]; i; i = ne[i]) {
            f[x] |= f[ver[i]];
        }
    }

    for (int i = 1; i <= n; i++) printf("%d\n", (int)f[i].count());
}

int main() {
    freopen("input.txt", "r", stdin);
    cin >> n >> m;

    // get edge
    memset(head, 0, sizeof head);
    for (int i = 0; i < m; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        add(x, y);
    }

    toposort();
    solve();
}