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
#include <bits/stdc++.h>
using namespace std;
const int N = 100000;
const long long MOD = 1000000007;
int n, p[N], x[N];
vector<int> graph[N];
long long colored[N], not_colored[N];
void dfs(int u) {
  if (x[u] == 0) {
    long long haveNone = 1;
    long long haveOne = 0;
    not_colored[u] = 1;
    for (int ed = 0; ed < (int)graph[u].size(); ed++) {
      int v = graph[u][ed];
      dfs(v);
      if (x[v] == 0) {
        haveOne = (haveOne * (not_colored[v] + colored[v]) % MOD) % MOD +
                  (haveNone * colored[v]) % MOD;
        haveOne = haveOne % MOD;
        haveNone = haveNone * ((not_colored[v] + colored[v]) % MOD) % MOD;
        haveNone = haveNone % MOD;
      } else {
        haveOne = (haveOne * colored[v]) % MOD + (haveNone * colored[v]) % MOD;
        haveOne = haveOne % MOD;
        haveNone = haveNone * ((not_colored[v] + colored[v]) % MOD) % MOD;
        haveNone = haveNone % MOD;
      }
    }
    colored[u] = haveOne;
    not_colored[u] = haveNone;
  } else {
    not_colored[u] = 0;
    colored[u] = 1;
    for (int ed = 0; ed < (int)graph[u].size(); ed++) {
      int v = graph[u][ed];
      dfs(v);
      if (x[v] == 0) {
        colored[u] = colored[u] * ((not_colored[v] + colored[v]) % MOD) % MOD;
        colored[u] = colored[u] % MOD;
      } else {
        colored[u] = colored[u] * colored[v] % MOD;
        colored[u] = colored[u] % MOD;
      }
    }
  }
}
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d", &p[i]);
    graph[p[i]].push_back(i);
  }
  p[0] = -1;
  for (int i = 0; i < n; i++) scanf("%d", &x[i]);
  dfs(0);
  printf("%I64d\n", colored[0]);
  return 0;
}
