codeforces396C

On Changing Tree

 CodeForces - 396C 

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 vat 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, ... p_n (1 ≤ p_i < i), where p_i 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 qlines 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

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

Output

2
1

Note

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

 

给出一棵以1为根的树，形式是从节点2开始给出每个节点的父亲节点；
然后是m次操作，操作分为两种，1 v， x， k，表示在以v为根的子树上添加，添加的法则是看这个节点与v节点的距离为i的话，加上x-i*k；2 v查询节点v的值。

sol：把dfs序搞出来了以后，区间操作就是子树操作，然后两个都是经典操作，对于1操作，先加上Depth[v]*k,再对查询的点p减去Depth[p]*k

 

/*
给出一棵以1为根的树，形式是从节点2开始给出每个节点的父亲节点；
然后是m次操作，操作分为两种，1 v， x， k，表示在以v为根的子树上添加，
添加的法则是看这个节点与v节点的距离为i的话，加上x-i*k；2 v查询节点v的值。
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=300005,M=600005;
const ll Mod=1000000007;
int n,Q;
inline void Ad(ll &x,ll y)
{
    x+=y; x-=(x>=Mod)?Mod:0; x+=(x<0)?Mod:0;
}
namespace Tree
{
    int tot=0,Next[M],to[M],head[N];
    inline void add(int x,int y)
    {
        Next[++tot]=head[x];
        to[tot]=y;
        head[x]=tot;
    }
    int In[N],Out[N],cnt=0,Depth[N];
    inline void dfs(int x)
    {
        int i;
        In[x]=++cnt;
        for(i=head[x];i;i=Next[i]) Depth[to[i]]=Depth[x]+1,dfs(to[i]);
        Out[x]=++cnt;
    }
    struct segment
    {
        ll S[N<<1];
        #define lowbit(x) ((x)&(-x))
        inline void Ins(int x,int Val)
        {
            while(x<=cnt)
            {
                Ad(S[x],Val); x+=lowbit(x);
            }
        }
        inline int Que(int x)
        {
            ll ans=0;
            while(x>0)
            {
                Ad(ans,S[x]); x-=lowbit(x);
            }
            return ans;
        }
    }SGT[2];
    inline void Solve()
    {
        int i;
        Depth[1]=0; dfs(1);
        R(Q);
        while(Q--)
        {
            ll opt,rt,Val,Del;
            R(opt); R(rt);
            if(opt==1)
            {
                Val=read()%Mod; R(Del);
                SGT[0].Ins(In[rt],(Val+Del*Depth[rt]%Mod)%Mod);
                SGT[0].Ins(Out[rt]+1,(-1)*(Val+Del*Depth[rt]%Mod)%Mod);
                SGT[1].Ins(In[rt],Del);
                SGT[1].Ins(Out[rt]+1,(-1)*Del);
            }
            else
            {
                ll tmp,oo;
                tmp=SGT[0].Que(In[rt]);
                oo=SGT[1].Que(In[rt]);
                Ad(tmp,(-1)*oo*Depth[rt]%Mod);
                Wl(tmp);
            }
        }
    }
}
#define T Tree
int main()
{
    int i;
    R(n);
    for(i=2;i<=n;i++)
    {
        int x=read(); T::add(x,i);
    }
    T::Solve();
    return 0;
}
/*
Input
3
1 1
3
1 1 2 1
2 1
2 2
Output
2
1

Input
10
1 2 3 4 4 3 3 6 7
10
1 6 13 98
1 7 17 66
1 5 32 39
1 1 9 5
1 7 27 11
1 1 24 79
1 5 87 86
2 2
1 5 9 38
2 5
Output
999999956
999999832
*/