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 = (int)1e5 + 10;
;
const int MOD = (int)1e9 + 7;
;
vector<long long> num[maxn];
long long dpb[maxn], dpw[maxn], n;
bool isb[maxn];
long long po(long long a, long long b) {
  if (!b) return 1;
  long long t = po(a, b / 2);
  t = (t * t) % MOD;
  if (b & 1) t = (t * a) % MOD;
  return t;
}
void dfs(int v, int pr) {
  for (int i = 0; i < num[v].size(); i++)
    if (num[v][i] != pr) dfs(num[v][i], v);
  if (isb[v]) {
    dpw[v] = 0;
    dpb[v] = 1;
    for (int i = 0; i < num[v].size(); i++)
      if (num[v][i] != pr) {
        dpb[v] *= dpb[num[v][i]] + dpw[num[v][i]];
        dpb[v] %= MOD;
      }
  } else {
    dpw[v] = 1;
    for (int i = 0; i < num[v].size(); i++)
      if (num[v][i] != pr) {
        dpw[v] *= dpb[num[v][i]] + dpw[num[v][i]];
        dpw[v] %= MOD;
      }
    dpb[v] = 0;
    for (int i = 0; i < num[v].size(); i++)
      if (num[v][i] != pr) {
        dpb[v] += (dpw[v] * po(dpb[num[v][i]] + dpw[num[v][i]], MOD - 2)) %
                  MOD * dpb[num[v][i]];
        dpb[v] %= MOD;
      }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    long long t;
    cin >> t;
    num[i + 1].push_back(t);
    num[t].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) cin >> isb[i];
  dfs(0, -1);
  cout << dpb[0];
}
