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
        
         #include
         
          #include
          
           #include
           
            using namespace std;typedef long long ll;const int N=1e5+5,MOD=1e9+7;inline int read(){    char c=getchar();int x=0,f=1;    while(c<'0'||c>'9'){
      if(c=='-')f=-1; c=getchar();}    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}    return x*f;}struct edge{    int v,ne;}e[N<<1];int cnt=0,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][2];void dp(int u,int fa){    if(w[u]) f[u][1]=1;    else f[u][0]=1;    for(int i=h[u];i;i=e[i].ne){        int v=e[i].v;        if(v==fa) continue;        dp(v,u);        f[u][1]=(f[u][1]*(f[v][0]+f[v][1])+f[u][0]*f[v][1])%MOD;        f[u][0]=f[u][0]*(f[v][0]+f[v][1])%MOD;    }}int main(){    n=read();    for(int i=1;i<=n-1;i++) ins(read(),i);    for(int i=0;i