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 MaxN = 1000005;
const double eps = 1e-8;
const double DINF = 1e100;
const int INF = 1000000006;
const long long LINF = 1000000000000000005ll;
const int mod = (int)1e9 + 7;
int n;
int color[MaxN];
vector<int> edges[MaxN];
long long f[MaxN][2];
long long tmp[MaxN][2];
bool vis[MaxN];
void dfs(int u) {
  vector<int> sons;
  for (auto v : edges[u]) {
    if (vis[v]) continue;
    vis[v] = 1;
    dfs(v);
    sons.push_back(v);
  }
  for (int i = 0; i <= sons.size(); ++i) tmp[i][0] = tmp[i][1] = 0;
  if (color[u] == 1)
    tmp[0][1] = 1;
  else
    tmp[0][0] = 1;
  for (int i = 0; i < sons.size(); ++i) {
    int v = sons[i];
    tmp[i + 1][1] = (tmp[i + 1][1] + tmp[i][0] * f[v][1] % mod) % mod;
    tmp[i + 1][1] =
        (tmp[i + 1][1] + tmp[i][1] * (f[v][0] + f[v][1]) % mod) % mod;
    tmp[i + 1][0] =
        (tmp[i + 1][0] + tmp[i][0] * (f[v][0] + f[v][1]) % mod) % mod;
  }
  f[u][0] = tmp[sons.size()][0];
  f[u][1] = tmp[sons.size()][1];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int x;
    scanf("%d", &x);
    edges[x].push_back(i);
    edges[i].push_back(x);
  }
  for (int i = 0; i < n; ++i) scanf("%d", color + i);
  vis[0] = 1;
  dfs(0);
  printf("%d\n", (int)f[0][1]);
  return 0;
}
