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 MOD = 1e9 + 7;
const int N = 1e5;
vector<int> g[N];
int dp[N][2], col[N], l[N], r[N];
void dfs(int v) {
  int sz = g[v].size();
  for (int i = 0; i < sz; i++) dfs(g[v][i]);
  if (col[v]) {
    int add = 1;
    for (int i = 0; i < sz; i++)
      add = (add * 1ll * (dp[g[v][i]][0] + dp[g[v][i]][1])) % MOD;
    dp[v][1] = (dp[v][1] + add) % MOD;
  } else {
    int add = 1;
    for (int i = 0; i < sz; i++)
      add = (add * 1ll * (dp[g[v][i]][0] + dp[g[v][i]][1])) % MOD;
    dp[v][0] = (dp[v][0] + add) % MOD;
    if (sz) {
      l[0] = (dp[g[v][0]][0] + dp[g[v][0]][1]) % MOD;
      r[sz - 1] = (dp[g[v][sz - 1]][0] + dp[g[v][sz - 1]][1]) % MOD;
    }
    for (int i = 1; i < sz; i++)
      l[i] = (l[i - 1] * 1ll * (dp[g[v][i]][0] + dp[g[v][i]][1])) % MOD;
    for (int i = sz - 2; i >= 0; i--)
      r[i] = (r[i + 1] * 1ll * (dp[g[v][i]][0] + dp[g[v][i]][1])) % MOD;
    for (int i = 0; i < sz; i++) {
      add = 1;
      add =
          (add * 1ll * (i ? l[i - 1] : 1) * (i != sz - 1 ? r[i + 1] : 1)) % MOD;
      add = (add * 1ll * dp[g[v][i]][1]) % MOD;
      dp[v][1] = (dp[v][1] + add) % MOD;
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int p;
    cin >> p;
    g[p].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> col[i];
  dfs(0);
  cout << dp[0][1] << endl;
  return 0;
}
