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>
const int mod = 1e9 + 7;
int head[100005], e, n;
long long dp[100005][2];
struct E {
  int to, next;
} edge[100005];
inline void addedge(int u, int v) {
  edge[e].to = v;
  edge[e].next = head[u];
  head[u] = e++;
}
void dfs(int u) {
  int i;
  for (i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    dfs(v);
    dp[u][1] =
        (dp[u][1] * (dp[v][1] + dp[v][0]) % mod + dp[u][0] * dp[v][1] % mod) %
        mod;
    dp[u][0] = dp[u][0] * (dp[v][1] + dp[v][0]) % mod;
  }
}
int main() {
  scanf("%d", &n);
  int i, tmp;
  e = 1;
  for (i = 1; i < n; i++) scanf("%d", &tmp), addedge(tmp, i);
  for (i = 0; i < n; i++) scanf("%d", &tmp), dp[i][tmp] = 1;
  dfs(0);
  printf("%I64d\n", dp[0][1]);
  return 0;
}
