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 = 100010;
const long long int mod = 1000000007LL;
int n;
struct Edge {
  int to, next;
} edge[3 * maxn];
int Adj[maxn], Size;
void init() {
  Size = 0;
  memset(Adj, -1, sizeof(Adj));
}
void Add_Edge(int u, int v) {
  edge[Size].to = v;
  edge[Size].next = Adj[u];
  Adj[u] = Size++;
}
int c[maxn];
long long int dp[maxn][2];
void dfs(int u, int fa) {
  dp[u][c[u]] = 1;
  for (int i = Adj[u]; ~i; i = edge[i].next) {
    int v = edge[i].to;
    if (v == fa) continue;
    dfs(v, u);
    dp[u][1] = (dp[u][1] * dp[v][1] % mod + dp[u][0] * dp[v][1] % mod +
                dp[u][1] * dp[v][0] % mod) %
               mod;
    dp[u][0] = (dp[u][0] * dp[v][0] % mod + dp[u][0] * dp[v][1] % mod) % mod;
  }
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int v;
    scanf("%d", &v);
    Add_Edge(i, v);
    Add_Edge(v, i);
  }
  for (int i = 0; i < n; i++) scanf("%d", c + i);
  dfs(0, 0);
  printf("%I64d\n", dp[0][1]);
  return 0;
}
