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;
long long M = 1e9 + 7;
const long long N = 1e5 + 7;
long long n;
vector<long long> kid[N];
bool col[N];
deque<long long> muls[N], mulr[N];
long long dp[N][2];
void solve(long long v) {
  if (!kid[v].size()) {
    if (col[v])
      dp[v][1] = 1;
    else
      dp[v][0] = 1;
    return;
  }
  for (auto u : kid[v]) solve(u);
  for (long long i = 0; i < kid[v].size(); i++) {
    long long u = kid[v][i];
    long long s = dp[u][0] + dp[u][1];
    if (i == 0)
      muls[v].push_back(s);
    else
      muls[v].push_back((s * muls[v].back()) % M);
  }
  for (long long i = kid[v].size() - 1; i >= 0; i--) {
    long long u = kid[v][i];
    long long s = dp[u][0] + dp[u][1];
    if (i == kid[v].size() - 1)
      mulr[v].push_front(s);
    else
      mulr[v].push_front((s * mulr[v][0]) % M);
  }
  if (col[v]) {
    dp[v][1] = mulr[v][0];
    return;
  }
  dp[v][0] = mulr[v][0];
  for (long long i = 0; i < kid[v].size(); i++) {
    long long u = kid[v][i];
    long long s = dp[u][1];
    if (i) s = (s * muls[v][i - 1]) % M;
    if (i + 1 < kid[v].size()) s = (s * mulr[v][i + 1]) % M;
    dp[v][1] = (dp[v][1] + s) % M;
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 2; i <= n; i++) {
    long long p;
    cin >> p;
    p++;
    kid[p].push_back(i);
  }
  for (long long i = 1; i <= n; i++) cin >> col[i];
  solve(1);
  cout << dp[1][1];
}
