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 long long int mod = 1e9 + 7;
const long long int mod2 = 1e9 + 9;
const long long int maxn = 1e6 + 10;
const long long inf = 1e18 + 18;
double pie = 3.1415926535;
long long int col[maxn], dp[maxn][2];
vector<long long int> a[maxn];
void dfs(long long int s, long long int par) {
  dp[s][0] = 1 - col[s];
  dp[s][1] = col[s];
  long long int i, old0, old1, node;
  for (i = 0; i < int(a[s].size()); i++) {
    node = a[s][i];
    if (node == par) continue;
    old0 = dp[s][0];
    old1 = dp[s][1];
    dp[s][0] = dp[s][1] = 0;
    dfs(node, s);
    dp[s][1] += ((dp[node][1] * old0) % mod + (dp[node][0] * old1) % mod) % mod;
    dp[s][0] += old0 * dp[node][0];
    dp[s][0] %= mod;
    dp[s][1] += dp[node][1] * old1;
    dp[s][1] %= mod;
    dp[s][0] += dp[node][1] * old0;
    dp[s][0] %= mod;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, x, i;
  cin >> n;
  for (i = 2; i <= n; i++) {
    cin >> x;
    a[x + 1].push_back(i);
    a[i].push_back(x + 1);
  }
  for (i = 1; i <= n; i++) {
    cin >> col[i];
  }
  dfs(1, -1);
  cout << dp[1][1] << endl;
}
