Write a program which manipulates a weighted rooted tree $T$ with the following operations:

* $add(v,w)$: add $w$ to all edges from the root to node $u$

* $getSum(u)$: report the sum of weights of all edges from the root to node $u$



The given tree $T$ consists of $n$ nodes and every node has a unique ID from $0$ to $n-1$ respectively where ID of the root is $0$. Note that all weights are initialized to zero.

Constraints

* All the inputs are given in integers
* $ 2 \leq n \leq 100000 $
* $ c_j < c_{j+1} $   $( 1 \leq j \leq k-1 )$
* $ 2 \leq q \leq 200000 $
* $ 1 \leq u,v \leq n-1 $
* $ 1 \leq w \leq 10000 $

Input

The input is given in the following format.

$n$
$node_0$
$node_1$
$node_2$
$:$
$node_{n-1}$
$q$
$query_1$
$query_2$
$:$
$query_{q}$


The first line of the input includes an integer $n$, the number of nodes in the tree.

In the next $n$ lines,the information of node $i$ is given in the following format:


ki c1 c2 ... ck


$k_i$ is the number of children of node $i$, and $c_1$ $c_2$ ... $c_{k_i}$ are node IDs of 1st, ... $k$th child of node $i$.

In the next line, the number of queries $q$ is given. In the next $q$ lines, $i$th query is given in the following format:


0 v w


or


1 u


The first integer represents the type of queries.'0' denotes $add(v, w)$ and '1' denotes $getSum(u)$.

Output

For each $getSum$ query, print the sum in a line.

Examples

Input

6
2 1 2
2 3 5
0
0
0
1 4
7
1 1
0 3 10
1 2
0 4 20
1 3
0 5 40
1 4


Output

0
0
40
150


Input

4
1 1
1 2
1 3
0
6
0 3 1000
0 2 1000
0 1 1000
1 1
1 2
1 3


Output

3000
5000
6000
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<list>
#include<string>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<ctime>
using namespace std;

typedef long long ll;

bool debug = false;
const int NIL = -1;
const int INF = 1000000000;
const int NUM = 110010;
clock_t START, END;

int V, Q;

struct HeapNode {
	int d, u;
	bool operator < (const HeapNode& rhs) const {
		return d > rhs.d;
	}
};

struct Edge {
	int from, to, dist;
	Edge(int u, int v, int d) :from(u), to(v), dist(d) {}
};

struct Dijkstra {
	int n, m;
	vector<Edge> edges;
	vector<int> G[NUM];

	ll cnt, res;
	int dep[NUM], p[NUM], siz[NUM], son[NUM], top[NUM];
	ll id[NUM];
	ll sumv[NUM << 2], addv[NUM << 2];

	void init(int n) {
		this->n = n;
		cnt = 0;
		dfs1(0, -1, 0);
		dfs2(0, 0);
	}

	void AddEdge(int from, int to, int dist) {
		edges.push_back(Edge(from, to, dist));
		m = edges.size();
		G[from].push_back(m - 1);
	}

	void maintain(int o, ll L, ll R) {
		sumv[o] = 0;
		if (R > L) {
			sumv[o] = sumv[o * 2] + sumv[o * 2 + 1];
		}
		sumv[o] += addv[o] * (R - L + 1);
	}

	void query(int o, ll L, ll R, ll add, ll ql, ll qr) {
		if (ql <= L && qr >= R) {
			res += sumv[o] + add * (R - L + 1);
		}
		else {
			ll M = L + (R - L) / 2;
			if (ql <= M)
				query(o * 2, L, M, add + addv[o], ql, qr);
			if (qr > M)
				query(o * 2 + 1, M + 1, R, add + addv[o], ql, qr);
		}
	}

	void update(int o, ll L, ll R, ll ql, ll qr, ll w) {
		ll M = L + (R - L) / 2;
		if (ql <= L && qr >= R)
			addv[o] += w;
		else {
			if (ql <= M)
				update(o * 2, L, M, ql, qr, w);
			if (qr > M)
				update(o * 2 + 1, M + 1, R, ql, qr, w);
		}
		maintain(o, L, R);
	}

	void dfs1(int u, int fa, int depth) {
		dep[u] = depth;
		p[u] = fa;
		siz[u] = 1;
		int maxson = -1;
		for (int i = 0; i < G[u].size(); i++) {
			Edge& e = edges[G[u][i]];
			int v = e.to;
			if (v == fa)
				continue;
			dfs1(v, u, depth + 1);
			siz[u] += siz[v];
			if (siz[v] > maxson) {
				son[u] = v;
				maxson = siz[v];
			}
		}
	}

	void dfs2(int u, int topf) {
		id[u] = ++cnt;
		top[u] = topf;
		if (!son[u])
			return;
		dfs2(son[u], topf);
		for (int i = 0; i < G[u].size(); i++) {
			Edge& e = edges[G[u][i]];
			int v = e.to;
			if (v == p[u] || v == son[u])
				continue;
			dfs2(v, v);
		}
	}

	ll qRange(int u) {
		ll ans = 0;
		while (top[u] != top[0]) {
			res = 0;
			query(1, 1, n, 0, id[top[u]], id[u]);
			ans += res;
			u = p[top[u]];
		}
		res = 0;
		query(1, 1, n, 0, id[0], id[u]);
		ans += res;
		return ans;
	}

	void qupdate(int v, ll w) {
		while (top[v] != top[0]) {
			update(1, 1, n, id[top[v]], id[v], w);
			v = p[top[v]];
		}
		update(1, 1, n, id[son[0]], id[v], w);
	}
};

Dijkstra solve;

int main(void)
{
	if (debug) {
		START = clock();
		freopen("in17.txt", "r", stdin);
		freopen("out.txt", "w", stdout);
	}
	int k, c, cmd, v, w, u;
	cin >> V;
	for (int i = 0; i < V; i++) {
		scanf("%d", &k);
		for (int j = 0; j < k; j++) {
			scanf("%d", &c);
			solve.AddEdge(i, c, 1);
			solve.AddEdge(c, i, 1);
		}
	}
	solve.init(V);
	cin >> Q;
	for (int i = 0; i < Q; i++) {
		scanf("%d", &cmd);
		if (cmd) {
			scanf("%d", &u);
			printf("%lld\n", solve.qRange(u));
		}
		else {
			scanf("%d%d", &v, &w);
			solve.qupdate(v, static_cast<ll>(w));
		}
	}
	if (debug) {
		END = clock();
		double endtime = (double)(END - START) / 1000;
		printf("total time = %lf s", endtime);
	}
	return 0;
}
