// WARN: Input should be a simple graph
//
// Returns: (BCC count, indices of BCC of each vertex)
//
// BCC index starts from 1
pair<int, vector<int>> ebcc(const vector<vector<int>>& ch) {
    int n = ch.size() - 1;
    vector<int> c(n + 1);

    vector<int> low(n + 1), dfn(n + 1);
    vector<bool> vis(n + 1);
    vector<int> stack;

    int time = 0;
    int cnt = 0;

    auto dfs = [&] (auto dfs, int v, int pa) -> void {
        low[v] = dfn[v] = ++time;
        stack.emplace_back(v);
        vis[v] = 1;
        for (auto&& u : ch[v]) {
            if (u == pa) continue;
            if (not dfn[u]) {
                dfs(dfs, u, v);
                chmin(low[v], low[u]);
            } else if (vis[u]) {
                chmin(low[v], dfn[u]);
            }
        }
        if (dfn[v] == low[v]) {
            cnt += 1;
            c[v] = cnt;
            while (stack.back() != v) {
                c[stack.back()] = cnt;
                vis[stack.back()] = 0;
                stack.pop_back();
            }
            stack.pop_back();
        }
    };

    for (int i = 1; i <= n; ++i) {
        if (not dfn[i]) {
            dfs(dfs, i, 0);
        }
    }
    return { cnt, c };
}


// INFO: This function treats vertex pairs that have multiple edges as eBCCs.
//
// Returns: (BCC count, indices of BCC of each vertex)
//
// BCC index starts from 1
pair<int, vector<int>> ebcc_with_mult(const vector<vector<pii>>& ch) {
    int n = ch.size() - 1;
    vector<int> c(n + 1);

    vector<int> low(n + 1), dfn(n + 1);
    vector<bool> vis(n + 1);
    vector<int> stack;

    int time = 0;
    int cnt = 0;

    auto dfs = [&] (auto dfs, int v, int prev_idx) -> void {
        low[v] = dfn[v] = ++time;
        stack.emplace_back(v);
        vis[v] = 1;
        for (auto&& [u, i] : ch[v]) {
            if (i == prev_idx) continue;
            if (not dfn[u]) {
                dfs(dfs, u, i);
                chmin(low[v], low[u]);
            } else if (vis[u]) {
                chmin(low[v], dfn[u]);
            }
        }
        if (dfn[v] == low[v]) {
            cnt += 1;
            c[v] = cnt;
            while (stack.back() != v) {
                c[stack.back()] = cnt;
                vis[stack.back()] = 0;
                stack.pop_back();
            }
            stack.pop_back();
        }
    };

    for (int i = 1; i <= n; ++i) {
        if (not dfn[i]) {
            dfs(dfs, i, -1);
        }
    }
    return { cnt, c };
}