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;
using ll = long long;
using pll = pair<ll, ll>;
using dd = long double;
namespace {
static bool constexpr dbg = 0;
ll constexpr N = 1.1e5, MO = 1e9 + 7;
ll n, d[N][2];
bool bl[N];
vector<ll> adj[N];
void init() {
  cin >> n;
  for (ll i = 1; i <= (ll)(n - 1); ++i) {
    ll p;
    cin >> p;
    adj[i].push_back(p);
    adj[p].push_back(i);
  }
  for (ll i = 0; i < (ll)(n); ++i) cin >> bl[i];
}
void dfs(ll c, ll p) {
  d[c][0] = 1;
  d[c][1] = 0;
  for (auto ch : adj[c])
    if (ch != p) {
      dfs(ch, c);
      d[c][1] = (d[c][1] * d[ch][0] + d[c][0] * d[ch][1]) % MO;
      d[c][0] = (d[c][0] * d[ch][0]) % MO;
    }
  if (bl[c])
    d[c][1] = d[c][0];
  else
    d[c][0] = (d[c][0] + d[c][1]) % MO;
}
void solve() {
  dfs(0, -1);
  cout << d[0][1];
}
}  // namespace
int main() {
  if (!dbg) {
    std::ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  cout << fixed << setprecision(20);
  init();
  solve();
  return 0;
}
