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;
int abs(int x) {
  if (x < 0) return -x;
  return x;
}
int addmod(int v1, int v2) {
  int v3 = v1 + v2;
  if (v3 >= 1000000007) v3 -= 1000000007;
  return v3;
}
long long dp[100005][2];
long long old[100005][2];
int color[100005];
vector<int> vv[100005];
void dfs(int pos) {
  dp[pos][0] = !color[pos];
  dp[pos][1] = color[pos];
  for (int i = (0); i < (vv[pos].size()); i++) {
    dfs(vv[pos][i]);
    old[pos][0] = dp[pos][0];
    old[pos][1] = dp[pos][1];
    dp[pos][1] = dp[pos][0] = 0;
    int v, u;
    v = pos;
    u = vv[pos][i];
    dp[v][0] += (old[v][0] * dp[u][1]) % 1000000007;
    dp[v][1] += (old[v][1] * dp[u][1]) % 1000000007;
    dp[v][1] += (old[v][1] * dp[u][0]) % 1000000007;
    dp[v][1] += (old[v][0] * dp[u][1]) % 1000000007;
    dp[v][0] += (old[v][0] * dp[u][0]) % 1000000007;
    dp[v][0] %= 1000000007;
    dp[v][1] %= 1000000007;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int val;
  for (int i = (0); i < (n - 1); i++) {
    scanf("%d", &val);
    vv[val].push_back(i + 1);
  }
  for (int i = (0); i < (n); i++) scanf("%d", &color[i]);
  dfs(0);
  cout << (dp[0][1]), cout << (endl);
}
