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 = 1e5 + 5;
int n, dp[MAXN][2], mod = 1e9 + 7;
bool A[MAXN];
vector<int> g[MAXN];
int plu(int a, int b) { return (a + b) % mod; }
int mul(int a, int b) { return ((long long)a * b) % (long long)mod; }
void dfs(int u, int p) {
  dp[u][0] = 1, dp[u][1] = 0;
  for (auto v : g[u])
    if (v != p) {
      dfs(v, u);
      dp[u][1] = plu(mul(dp[u][1], dp[v][0]), mul(dp[u][0], dp[v][1]));
      dp[u][0] = mul(dp[u][0], dp[v][0]);
    }
  if (A[u])
    dp[u][1] = dp[u][0];
  else
    dp[u][0] = plu(dp[u][0], dp[u][1]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int v;
    scanf("%d", &v);
    g[i].emplace_back(v), g[v].emplace_back(i);
  }
  for (int i = 0; i < n; ++i) scanf("%d", A + i);
  dfs(0, -1);
  printf("%d\n", dp[0][1]);
}
