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 = 111111;
const int maxm = 111111;
const int mod = 1000000007;
int n;
vector<int> a[maxn];
int col[maxn];
int s[maxn];
int num;
int v[maxn];
long long dp[maxn][2];
void dfs(int x) {
  v[x] = 1;
  int len = a[x].size();
  for (int i = 0; i < len; i++)
    if (!v[a[x][i]]) {
      dfs(a[x][i]);
    }
  s[num++] = x;
}
void solve() {
  num = 0;
  int t;
  memset(v, 0, sizeof(v));
  for (int i = 0; i <= n; i++) {
    a[i].clear();
  }
  for (int i = 0; i < n - 1; i++) {
    scanf("%d", &t);
    a[i + 1].push_back(t);
    a[t].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &col[i]);
  }
  dfs(0);
  int len;
  memset(v, 0, sizeof(v));
  for (int i = 0; i < num; i++) {
    t = s[i];
    v[t] = 1;
    if (col[t] == 1) {
      dp[t][0] = 0;
      dp[t][1] = 1;
      len = a[t].size();
      for (int i = 0; i < len; i++)
        if (v[a[t][i]]) {
          dp[t][1] = dp[t][1] * (dp[a[t][i]][1] + dp[a[t][i]][0]) % mod;
        }
    } else {
      dp[t][0] = 1;
      dp[t][1] = 0;
      len = a[t].size();
      for (int i = 0; i < len; i++)
        if (v[a[t][i]]) {
          dp[t][1] = (dp[t][1] * (dp[a[t][i]][1] + dp[a[t][i]][0]) +
                      dp[t][0] * dp[a[t][i]][1]) %
                     mod;
          dp[t][0] = dp[t][0] * (dp[a[t][i]][1] + dp[a[t][i]][0]) % mod;
        }
    }
  }
  cout << dp[0][1] << endl;
}
int main() {
  int t;
  while (~scanf("%d", &n)) {
    solve();
  }
  return 0;
}
