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 <bits/stdc++.h>
#define pb push_back
using namespace std;
typedef long long ll;
const int maxn=100005;

int n,q,cnt;
vector<int>G[maxn];
int fa[maxn],dep[maxn],size[maxn],son[maxn];
int top[maxn],id[maxn],rk[maxn];
ll seg[4*maxn],lazy[4*maxn];

void dfs1(int u,int p) {
    fa[u]=p;
    dep[u]=dep[p]+1;
    size[u]=1;
    for (int v:G[u]) {
        if (v==p) continue; 
        dfs1(v,u);
        size[u]+=size[v];
        if (size[son[u]]<size[v])
            son[u]=v;
    }
}

void dfs2(int u,int t) {
    top[u]=t;
    id[u]=++cnt;
    rk[cnt]=u;
    if (son[u]) dfs2(son[u],t);
    for (int v:G[u])
        if (v!=son[u] && v!=fa[u])
            dfs2(v,v);
}

void add(int id,int l,int r,int ql,int qr,int v) {
    if (lazy[id]) {
        seg[id]+=lazy[id]*(r-l+1);
        if (l!=r) {
            lazy[id<<1]+=lazy[id];
            lazy[id<<1|1]+=lazy[id];
        }
        lazy[id]=0;
    }
    if (ql==l && qr==r) {
        lazy[id]+=v;
        return;
    }
    int m=l+r>>1;
    if (qr<=m) add(id<<1,l,m,ql,qr,v);
    else if (ql>m) add(id<<1|1,m+1,r,ql,qr,v);
    else {
        add(id<<1,l,m,ql,m,v);
        add(id<<1|1,m+1,r,m+1,qr,v);
    }
    seg[id]=seg[id<<1]+seg[id<<1|1]+lazy[id<<1]*(m-l+1)+lazy[id<<1|1]*(r-m);
}

ll query(int id,int l,int r,int ql,int qr) {
    if (lazy[id]) {
        seg[id]+=lazy[id]*(r-l+1);
        if (l!=r) {
            lazy[id<<1]+=lazy[id];
            lazy[id<<1|1]+=lazy[id];
        }
        lazy[id]=0;
    }
    if (ql==l && qr==r) return seg[id];
    int m=l+r>>1;
    if (qr<=m) return query(id<<1,l,m,ql,qr);
    else if (ql>m) return query(id<<1|1,m+1,r,ql,qr);
    else return query(id<<1,l,m,ql,m)+query(id<<1|1,m+1,r,m+1,qr);
}

void update(int v,int w) {
    while (top[1]!=top[v]) {
        add(1,1,n,id[top[v]],id[v],w);
        v=fa[top[v]];
    }
    add(1,1,n,id[1],id[v],w);
}

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

int main() {
    cin>>n;
    for (int i=1;i<=n;i++) {
        int m; cin>>m;
        for (int j=0;j<m;j++) {
            int v; cin>>v; ++v;
            G[i].pb(v);
        }
    }
    dfs1(1,0);
    dfs2(1,1);
    cin>>q;
    for (int i=0;i<q;i++) {
        bool op; cin>>op;
        if (op) {
            int u; cin>>u; ++u;
            cout<<getsum(u)-getsum(1)<<'\n';
        }
        else {
            int v,w; cin>>v>>w; ++v;
            update(v,w);
        }
    }
}

