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 = 100001;
int n, k;
vector<vector<int> > adjList(N);
vector<int> nxt(N, -1);
bool color[N];
int dp[N][2][2];
void dfs(int node) {
  bool leaf = true;
  for (int i = 0; i < adjList[node].size(); i++) {
    leaf = false;
    int ch = adjList[node][i];
    int nextCh = i == adjList[node].size() - 1 ? -1 : adjList[node][i + 1];
    nxt[ch] = nextCh;
    dfs(ch);
  }
  if (leaf) adjList[node].push_back(-1);
}
long long solve(int node, bool root, bool black) {
  if (node == -1) return black ? 0 : 1;
  if (dp[node][root][black] != -1) return dp[node][root][black];
  long long ans = 0;
  if (root) {
    ans = solve(adjList[node][0], false, !color[node]);
  } else {
    ans = solve(node, true, true) * solve(nxt[node], root, black) % 1000000007;
    if (black) {
      if (color[node]) {
        ans = (ans + solve(adjList[node][0], false, false) *
                         solve(nxt[node], false, false) % 1000000007) %
              1000000007;
      } else {
        ans = (ans + solve(adjList[node][0], false, true) *
                         solve(nxt[node], false, false) % 1000000007) %
              1000000007;
        ans = (ans + solve(adjList[node][0], false, false) *
                         solve(nxt[node], false, true) % 1000000007) %
              1000000007;
      }
    } else {
      if (!color[node]) {
        ans = (ans + solve(adjList[node][0], false, false) *
                         solve(nxt[node], false, false) % 1000000007) %
              1000000007;
      }
    }
  }
  return dp[node][root][black] = ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x;
    scanf("%d", &x);
    adjList[x].push_back(i);
  }
  for (int i = 0; i < n; i++) scanf("%d", &color[i]);
  dfs(0);
  memset(dp, -1, sizeof dp);
  printf("%d", solve(0, true, true));
  return 0;
}
