On Changing Tree

C. On Changing Tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a rooted tree consisting of n vertices numbered from 1 to n. The root of the tree is a vertex number 1.

Initially all vertices contain number 0. Then come q queries, each query has one of the two types:

• The format of the query: 1 v x k. In response to the query, you need to add to the number at vertex v number x; to the numbers at the descendants of vertex v at distance 1, add x - k; and so on, to the numbers written in the descendants of vertex v at distance i, you need to add x - (i·k). The distance between two vertices is the number of edges in the shortest path between these vertices.
• The format of the query: 2 v. In reply to the query you should print the number written in vertex v modulo 1000000007 (109 + 7).

Process the queries given in the input.

Input

The first line contains integer n (1 ≤ n ≤ 3·105) — the number of vertices in the tree. The second line contains n - 1 integers p2, p3, ... pn(1 ≤ pi < i), where pi is the number of the vertex that is the parent of vertex i in the tree.

The third line contains integer q (1 ≤ q ≤ 3·105) — the number of queries. Next q lines contain the queries, one per line. The first number in the line is type. It represents the type of the query. If type = 1, then next follow space-separated integers v, x, k (1 ≤ v ≤ n; 0 ≤ x < 109 + 7; 0 ≤ k < 109 + 7). If type = 2, then next follows integer v (1 ≤ v ≤ n) — the vertex where you need to find the value of the number.

Output

For each query of the second type print on a single line the number written in the vertex from the query. Print the number modulo 1000000007 (109 + 7).

Examples

input

Copy

3
1 1
3
1 1 2 1
2 1
2 2

output

Copy

2
1

Note

You can read about a rooted tree here: http://en.wikipedia.org/wiki/Tree_(graph_theory).

#pragma GCC optimize(2)

#include <bits/stdc++.h>
#define lowbit(x) x&(-x)
using namespace std;
const int maxn = 1e6 + 108;
typedef long long ll;
const ll mod=1e9+7;

int n,q;
int fa[maxn],dfn[maxn],siz[maxn],dep[maxn];
vector<int>e[maxn];

inline int dfs(int cur){
    static int tot=0;
    dfn[cur]=++tot;
    for(register int i=0;i<e[cur].size();++i){
        int to=e[cur][i];
        siz[cur]+=dfs(to);
    }
    //printf("debug dfn[%d] = %d\n",cur,dfn[cur]);
    return ++siz[cur];
}

struct BIT{
    ll o[maxn];
    inline void init(){
        memset(o,0,sizeof(o));
    }
    inline void update(int pos,ll val){
        while(pos<=n){
            o[pos]=(o[pos]+val)%mod;
            pos+=lowbit(pos);
        }
    }

    inline ll query(int pos){
        ll res=0;
        while(pos){
            res=(res+o[pos])%mod;
            pos-=lowbit(pos);
        }
        return res%mod;
    }

}atom1,atom2;



int main() {
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    scanf("%d",&n);
    for(register int i=2;i<=n;++i){
        scanf("%d",&fa[i]);
        e[fa[i]].emplace_back(i);
        dep[i]=dep[fa[i]]+1;
    }
    atom1.init();
    atom2.init();
    dfs(1);
    scanf("%d",&q);
    int type,v,x;
    ll k;
    while(q--){
        scanf("%d",&type);
        if(type==1){
            scanf("%d%d%lld",&v,&x,&k);
            atom1.update(dfn[v],x+dep[v]*k%mod);
            atom1.update(dfn[v]+siz[v],mod-(x+dep[v]*k%mod)%mod);
            atom2.update(dfn[v],k);
            atom2.update(dfn[v]+siz[v],(mod-k%mod)%mod);
        }
        else{
            scanf("%d",&v);
            printf("%lld\n",(atom1.query(dfn[v])%mod-dep[v]*atom2.query(dfn[v])%mod+mod)%mod);
        }
    }
    return 0;
}