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 + 10, mod = 1e9 + 7;
int head[N], nxt[N << 1], to[N << 1], cnt;
int n, col[N];
long long dp[N][2];
void init() {
  memset(head, -1, sizeof(head));
  cnt = 0;
}
void addEdge(int u, int v) {
  nxt[cnt] = head[u];
  to[cnt] = v;
  head[u] = cnt++;
}
void dfs(int u, int pre) {
  dp[u][0] = 1;
  dp[u][1] = 0;
  for (int i = head[u]; ~i; i = nxt[i]) {
    int v = to[i];
    if (v == pre) continue;
    dfs(v, u);
    dp[u][1] = dp[u][1] * dp[v][0] % mod;
    dp[u][1] = (dp[u][1] + dp[u][0] * dp[v][1]) % mod;
    dp[u][0] = dp[u][0] * dp[v][0] % mod;
  }
  if (col[u])
    dp[u][1] = dp[u][0];
  else
    dp[u][0] = (dp[u][0] + dp[u][1]) % mod;
}
int main() {
  while (~scanf("%d", &n)) {
    init();
    for (int i = 0, j; i < n - 1; i++) {
      scanf("%d", &j);
      addEdge(j, i + 1);
      addEdge(i + 1, j);
    }
    for (int i = 0; i < n; i++) scanf("%d", col + i);
    dfs(0, -1);
    printf("%I64d\n", dp[0][1]);
  }
  return 0;
}
