Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;

#ifndef MAX
#define MAX 400000
#endif

//Range Add Query+Range Sum Query
template<class T>
class RAQ_RSQ {
public:
	int n;
	T dat[MAX], lazy[MAX], ZERO;
	void init(int n_) {
		ZERO = T();
		n = 1; while (n < n_)n <<= 1;
		for (int i = 0; i < 2 * n - 1; i++) {
			dat[i] = lazy[i] = ZERO;
		}
	}
	inline void push(int k, int s) {
		dat[k] = dat[k] + lazy[k] * s;
		if (k < n - 1) {
			lazy[k * 2 + 1] = lazy[k * 2 + 1] + lazy[k];
			lazy[k * 2 + 2] = lazy[k * 2 + 2] + lazy[k];
		}
		lazy[k] = ZERO;
	}
	inline void update_node(int k) {
		dat[k] = dat[k * 2 + 1] + dat[k * 2 + 2];
	}
	inline void update(int a, int b, T x, int k, int l, int r) {
		push(k, r - l);
		if (r <= a || b <= l)return;
		if (a <= l&&r <= b) {
			lazy[k] = lazy[k] + x; push(k, r - l); return;
		}
		update(a, b, x, k * 2 + 1, l, (l + r) / 2);
		update(a, b, x, k * 2 + 2, (l + r) / 2, r);
		update_node(k);
	}
	inline T query(int a, int b, int k, int l, int r) {
		push(k, r - l);
		if (r <= a || b <= l)return ZERO;
		if (a <= l&&r <= b)return dat[k];
		T lb = query(a, b, k * 2 + 1, l, (l + r) / 2);
		T rb = query(a, b, k * 2 + 2, (l + r) / 2, r);
		update_node(k);
		return lb + rb;
	}
	inline void update(int a, int b, T x) {
		update(a, b, x, 0, 0, n);
	}
	inline void update(int a, T x) {
		update(a, a + 1, x);
	}
	inline T query(int a, int b) {
		return query(a, b, 0, 0, n);
	}
	inline T query(int a) {
		return query(a, a + 1);
	}
};

vector<int>E[200000];
int par[200000],d[200000],sub[200000];
int chain[200000],vid[200000];
int node_num;

void dfs(int v,int p,int dep){
	par[v]=p;d[v]=dep;
	sub[v]=1;
	for(int u:E[v]){
		if(u==p)continue;
		dfs(u,v,dep+1);
		sub[v]+=sub[u];
	}
}
void hld(int v,int k){
	chain[v]=k;
	vid[v]=node_num++;
	int Max=0,id=-1;
	for(int u:E[v]){
		if(u==par[v])continue;
		if(Max<sub[u]){
			Max=sub[u];
			id=u;
		}
	}
	if(id==-1)return;
	hld(id,chain[v]);
	for(int u:E[v]){
		if(u==id||u==par[v])continue;
		hld(u,u);
	}
}
ll res=0;
void query(int u,int v,function<void(int,int)>f){
	while(1){
		if(d[chain[u]]>d[chain[v]])swap(u,v);
		if(chain[u]==chain[v]){
			f(min(vid[u],vid[v])+1,max(vid[u],vid[v])+1);
			break;
		}
		else{
			f(vid[chain[v]],vid[v]+1);
			v=par[chain[v]];
		}
	}
}

RAQ_RSQ<ll>seg;
int main(){
	int n;scanf("%d",&n);
	rep(i,n){
		int k;scanf("%d",&k);
		rep(j,k){
			int c;scanf("%d",&c);
			E[i].push_back(c);
		}
	}
	dfs(0,-1,0);
	hld(0,0);
	seg.init(n);
	int q;scanf("%d",&q);
	rep(i,q){
		int t;scanf("%d",&t);
		if(t==0){
			int v,w;scanf("%d%d",&v,&w);
			query(0,v,[&](int l,int r){seg.update(l,r,w);});
		}
		else{
			int u;scanf("%d",&u);
			res=0;
			query(0,u,[&](int l,int r){res+=seg.query(l,r);});
			printf("%lld\n",res);
		}
	}
}