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 = 1e5 + 11;
const int Mod = 1e9 + 7;
struct Edge {
  int next, to;
  Edge() {}
  Edge(int a, int b) : next(a), to(b) {}
};
int head[N];
Edge err[N << 1];
int nedge;
int isb[N];
long long dp[N][2];
int n;
void add_edge(int a, int b) {
  err[++nedge] = Edge(head[a], b);
  head[a] = nedge;
  err[++nedge] = Edge(head[b], a);
  head[b] = nedge;
}
long long inv(long long x, long long p) {
  long long ret = 1;
  for (int i = p - 2; i; i >>= 1, x = (x * x) % p) {
    if (i & 1) ret = (ret * x) % p;
  }
  return ret;
}
void dfs(int u, int fa) {
  if (isb[u])
    dp[u][0] = 0, dp[u][1] = 1;
  else
    dp[u][0] = 1, dp[u][1] = 0;
  long long ss = 1, xx = 0;
  for (int i = head[u]; i != -1; i = err[i].next) {
    int v = err[i].to;
    if (v != fa) {
      dfs(v, u);
      ss *= (dp[v][0] + dp[v][1]);
      ss %= Mod;
    }
  }
  for (int i = head[u]; i != -1; i = err[i].next) {
    int v = err[i].to;
    if (v != fa) {
      xx += ((ss * inv(dp[v][0] + dp[v][1], Mod)) % Mod * dp[v][1]) % Mod;
      xx %= Mod;
    }
  }
  if (isb[u])
    dp[u][0] = 0, dp[u][1] = ss;
  else
    dp[u][0] = ss, dp[u][1] = xx;
}
int main() {
  scanf("%d", &n);
  memset(head, -1, sizeof(head));
  nedge = -1;
  for (int i = 1; i < n; ++i) {
    int x;
    scanf("%d", &x);
    add_edge(i, x);
  }
  for (int i = 0; i < n; ++i) scanf("%d", &isb[i]);
  dfs(0, -1);
  printf("%lld\n", dp[0][1]);
}
