Codeforces 461B. Appleman and Tree[树形DP 方案数]

B. Appleman and Tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Appleman has a tree with n vertices. Some of the vertices (at least one) are colored black and other vertices are colored white.

Consider a set consisting of k (0 ≤ k < n) edges of Appleman's tree. If Appleman deletes these edges from the tree, then it will split into(k + 1) parts. Note, that each part will be a tree with colored vertices.

Now Appleman wonders, what is the number of sets splitting the tree in such a way that each resulting part will have exactly one black vertex? Find this number modulo 1000000007 (109 + 7).

Input

The first line contains an integer n (2  ≤ n ≤ 105) — the number of tree vertices.

The second line contains the description of the tree: n - 1 integers p0, p1, ..., pn - 2 (0 ≤ pi ≤ i). Where pi means that there is an edge connecting vertex (i + 1) of the tree and vertex pi. Consider tree vertices are numbered from 0 to n - 1.

The third line contains the description of the colors of the vertices: n integers x0, x1, ..., xn - 1 (xi is either 0 or 1). If xi is equal to 1, vertex i is colored black. Otherwise, vertex i is colored white.

Output

Output a single integer — the number of ways to split the tree modulo 1000000007 (109 + 7).

Examples

input

3
0 0
0 1 1

output

2

input

6
0 1 1 0 4
1 1 0 0 1 0

output

1

input

10
0 1 2 1 4 4 4 0 8
0 0 0 1 0 1 1 0 0 1

output

27

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题意：分成若干个连通块，每个只有一个黑色节点，求方案数

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f[i][0/1]表示以i为根的子树i是否在有黑色节点的连通块中的方案数

f[u][1]=(f[u][1]*(f[v][0]+f[v][1])+f[u][0]*f[v][1])%MOD; v是0 u跟他相连，v是1 不相连；u是0时要跟v是1相连
f[u][0]=f[u][0]*(f[v][0]+f[v][1])%MOD;同理

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int N=1e5+,MOD=1e9+;
inline int read(){
    char c=getchar();int x=,f=;
    while(c<''||c>''){if(c=='-')f=-; c=getchar();}
    while(c>=''&&c<=''){x=x*+c-''; c=getchar();}
    return x*f;
}
struct edge{
    int v,ne;
}e[N<<];
int cnt=,h[N],w[N];
inline void ins(int u,int v){
    cnt++;
    e[cnt].v=v;e[cnt].ne=h[u];h[u]=cnt;
    cnt++;
    e[cnt].v=u;e[cnt].ne=h[v];h[v]=cnt;
}
int n;ll f[N][];
void dp(int u,int fa){
    if(w[u]) f[u][]=;
    else f[u][]=;
    for(int i=h[u];i;i=e[i].ne){
        int v=e[i].v;
        if(v==fa) continue;
        dp(v,u);
        f[u][]=(f[u][]*(f[v][]+f[v][])+f[u][]*f[v][])%MOD;
        f[u][]=f[u][]*(f[v][]+f[v][])%MOD;
    }
}
int main(){
    n=read();
    for(int i=;i<=n-;i++) ins(read(),i);
    for(int i=;i<n;i++) w[i]=read();
    dp(,-);
    cout<<f[][];
}