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;
inline long long read() {
  char i = getchar();
  long long f = 1, res = 0;
  while (i < '0' || i > '9') {
    if (i == '-') f = -1;
    i = getchar();
  }
  while (i >= '0' && i <= '9') {
    res = res * 10 + i - '0';
    i = getchar();
  }
  return res * f;
}
const int N = 1e5 + 50;
const long long mod = 1e9 + 7;
int k, head[N];
long long a[N], dp[N][2];
struct zk {
  int v, next;
} ed[N << 1];
inline void adde(int u, int v) {
  ed[++k].v = v;
  ed[k].next = head[u];
  head[u] = k;
}
inline void dfs(int u, int fa) {
  for (register int i = head[u]; ~i; i = ed[i].next) {
    int v = ed[i].v;
    if (v == fa) continue;
    dfs(v, u);
    dp[u][1] =
        (dp[u][1] * (dp[v][0] + dp[v][1]) % mod + dp[u][0] * dp[v][1] % mod) %
        mod;
    dp[u][0] = (dp[u][0] * (dp[v][0] + dp[v][1])) % mod;
  }
}
int main() {
  memset(head, -1, sizeof head);
  int n = read();
  for (register int i = 2; i <= n; ++i) {
    int p = read() + 1;
    adde(i, p);
    adde(p, i);
  }
  for (register int i = 1; i <= n; ++i) {
    a[i] = read();
    dp[i][a[i]] = 1;
  }
  dfs(1, 0);
  printf("%lld", dp[1][1]);
}
