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 long long mod = 1000000007;
const int maxn = 110000;
vector<int> tree[maxn];
bool vis[maxn];
int color[maxn];
long long f[maxn][2];
long long t1, t2;
void dfs(int root) {
  vis[root] = true;
  int siz = tree[root].size();
  if (color[root] == 1) {
    f[root][1] = 1;
    f[root][0] = 0;
  } else {
    f[root][0] = 1;
    f[root][1] = 0;
  }
  for (int i = 0; i < siz; ++i) {
    int v = tree[root][i];
    if (!vis[tree[root][i]]) {
      dfs(tree[root][i]);
      if (color[root] == 1) {
        f[root][1] = f[root][1] * (f[v][0] + f[v][1]) % mod;
      } else {
        t1 = f[root][0] * (f[v][0] + f[v][1]);
        t2 = f[root][1] * (f[v][0] + f[v][1]) + f[root][0] * f[v][1];
        f[root][1] = t2 % mod;
        f[root][0] = t1 % mod;
      }
    }
  }
}
int main(int argc, char* argv[]) {
  int n, p;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d", &p);
    tree[i].push_back(p);
    tree[p].push_back(i);
  }
  for (int i = 0; i < n; ++i) scanf("%d", &color[i]);
  dfs(0);
  printf("%I64d\n", f[0][1]);
  return 0;
}
