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>
const double PI = 3.141592653589793238460;
const int dx4[] = {0, 0, -1, 1};
const int dy4[] = {1, -1, 0, 0};
const int dx8[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int dy8[] = {0, 0, -1, 1, -1, 1, -1, 1};
void input() {}
using namespace std;
vector<int> g[100005];
bool f[100005];
long long dp[100005][2];
void dfs(int v, int par) {
  dp[v][f[v]] = 1;
  for (auto ch : g[v]) {
    if (ch != par) {
      int u = ch;
      dfs(ch, v);
      long long x0 = dp[v][0];
      long long x1 = dp[v][1];
      long long y0 = dp[u][0];
      long long y1 = dp[u][1];
      dp[v][0] = (x0 * y0 + x0 * y1) % 1000000007;
      dp[v][1] = (x1 * y1 + x0 * y1 + x1 * y0) % 1000000007;
    }
  }
}
void solve1() {
  int n, i, a;
  cin >> n;
  for (int i = 0; i < n - 1; i++) cin >> a, g[a].push_back(i + 1);
  for (i = 0; i < n; i++) cin >> a, f[i] = a;
  dfs(0, -1);
  printf("%d\n", dp[0][1]);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  input();
  solve1();
  return 0;
}
