D. 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).

Sample test(s) 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

題意：給了一棵樹以及每個節點的顏色，1代表黑，0代表白，要求的是，如果將這棵樹拆成k棵樹，使得每棵樹恰好有一個黑色節點 思路：樹形dp           dp[u][0]表示以u為根的子樹對父親的貢獻為0           dp[u][1]表示以u為根的子樹對父親的貢獻為1           我現在假設u為白色，它的子樹有x，y，z，那麼有           dp[u][1]+=dp[x][1]*dp[y][0]*dp[z][0]+dp[x][0]*dp[y][1]*dp[z][0]+dp[x][0]*dp[y][0]*dp[z][1]           dp[u][0]+=dp[x][0]*dp[y][0]*dp[z][0]           然後判斷u的顏色，假設u的父親為p          1.為黑色，不切斷邊（u，p），那麼dp[u][1]=dp[u][0]，切斷（u，p），那麼dp[u][0]不變          2.為白色，如果切斷（u，p），dp[u][0]還要加上dp[u][1]

#include 
#include 
#include 
#include 
using namespace std;
#define mod 1000000007
#define maxn 200010
typedef long long ll;
int vex[maxn];
int head[maxn];
int edge[maxn<<1];
int next[maxn];
ll dp[maxn][2];
int d;

void add(int u,int v)
{
    edge[d]=v;
    next[d]=head[u];
    head[u]=d++;
}

void addit(ll &a,ll x)
{
    a+=x;
    if(a>=mod)a-=mod;
}

void dfs(int u,int pre)
{
    int i,j,k;
    dp[u][0]=1;
    dp[u][1]=0;
    for(i=head[u];i!=-1;i=next[i]){
        int v=edge[i];
        if(v!=pre){
            dfs(v,u);
            dp[u][1]=dp[u][1] * dp[v][0] % mod;
            addit( dp[u][1] , dp[u][0] * dp[v][1] % mod);
            dp[u][0]=dp[u][0] * dp[v][0] % mod;
        }
    }
    if(vex[u])dp[u][1]=dp[u][0];
    else addit(dp[u][0],dp[u][1]);
}

int main()
{
    int n;

    while(scanf("%d",&n)!=EOF)
    {
        int i,j;
        d=0;
        memset(head,-1,sizeof(head));
        for(i=1;i