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 ld = long double;
using P = pair<ll, ll>;
constexpr ll inf = 1000000000;
constexpr ll mod = 1000000007;
constexpr long double eps = 1e-6;
template <typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> p) {
  os << to_string(p.first) << " " << to_string(p.second);
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
  for (ll i = (0); i < (ll)(v.size()); i++) {
    if (i) os << " ";
    os << v[i];
  }
  return os;
}
struct modint {
  ll n;

 public:
  modint(const ll n = 0) : n((n % mod + mod) % mod) {}
  static modint pow(modint a, int m) {
    modint r = 1;
    while (m > 0) {
      if (m & 1) {
        r *= a;
      }
      a = (a * a);
      m /= 2;
    }
    return r;
  }
  modint& operator++() {
    *this += 1;
    return *this;
  }
  modint& operator--() {
    *this -= 1;
    return *this;
  }
  modint operator++(int) {
    modint ret = *this;
    *this += 1;
    return ret;
  }
  modint operator--(int) {
    modint ret = *this;
    *this -= 1;
    return ret;
  }
  modint operator~() const { return (this->pow(n, mod - 2)); }
  friend bool operator==(const modint& lhs, const modint& rhs) {
    return lhs.n == rhs.n;
  }
  friend bool operator<(const modint& lhs, const modint& rhs) {
    return lhs.n < rhs.n;
  }
  friend bool operator>(const modint& lhs, const modint& rhs) {
    return lhs.n > rhs.n;
  }
  friend modint& operator+=(modint& lhs, const modint& rhs) {
    lhs.n += rhs.n;
    if (lhs.n >= mod) lhs.n -= mod;
    return lhs;
  }
  friend modint& operator-=(modint& lhs, const modint& rhs) {
    lhs.n -= rhs.n;
    if (lhs.n < 0) lhs.n += mod;
    return lhs;
  }
  friend modint& operator*=(modint& lhs, const modint& rhs) {
    lhs.n = (lhs.n * rhs.n) % mod;
    return lhs;
  }
  friend modint& operator/=(modint& lhs, const modint& rhs) {
    lhs.n = (lhs.n * (~rhs).n) % mod;
    return lhs;
  }
  friend modint operator+(const modint& lhs, const modint& rhs) {
    return modint(lhs.n + rhs.n);
  }
  friend modint operator-(const modint& lhs, const modint& rhs) {
    return modint(lhs.n - rhs.n);
  }
  friend modint operator*(const modint& lhs, const modint& rhs) {
    return modint(lhs.n * rhs.n);
  }
  friend modint operator/(const modint& lhs, const modint& rhs) {
    return modint(lhs.n * (~rhs).n);
  }
};
istream& operator>>(istream& is, modint m) {
  is >> m.n;
  return is;
}
ostream& operator<<(ostream& os, modint m) {
  os << m.n;
  return os;
}
long long extgcd(long long a, long long b, long long& x, long long& y) {
  long long d = a;
  if (b != 0) {
    d = extgcd(b, a % b, y, x);
    y -= (a / b) * x;
  } else {
    x = 1;
    y = 0;
  }
  return d;
}
long long mod_inverse(long long a, long long m) {
  long long x, y;
  if (extgcd(a, m, x, y) == 1)
    return (m + x % m) % m;
  else
    return -1;
}
vector<long long> fact(1010101 + 1, inf);
long long mod_fact(long long n, long long& e) {
  if (fact[0] == inf) {
    fact[0] = 1;
    if (1010101 != 0) fact[1] = 1;
    for (ll i = 2; i <= 1010101; ++i) {
      fact[i] = (fact[i - 1] * i) % mod;
    }
  }
  e = 0;
  if (n == 0) return 1;
  long long res = mod_fact(n / mod, e);
  e += n / mod;
  if ((n / mod) % 2 != 0) return (res * (mod - fact[n % mod])) % mod;
  return (res * fact[n % mod]) % mod;
}
long long mod_comb(long long n, long long k) {
  if (n < 0 || k < 0 || n < k) return 0;
  long long e1, e2, e3;
  long long a1 = mod_fact(n, e1), a2 = mod_fact(k, e2),
            a3 = mod_fact(n - k, e3);
  if (e1 > e2 + e3) return 0;
  return (a1 * mod_inverse((a2 * a3) % mod, mod)) % mod;
}
using mi = modint;
mi mod_pow(mi a, ll n) {
  mi ret = 1;
  mi tmp = a;
  while (n > 0) {
    if (n % 2) ret *= tmp;
    tmp = tmp * tmp;
    n /= 2;
  }
  return ret;
}
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int n;
vector<vector<int>> t;
vector<int> x;
pair<mi, mi> dfs(int v, int p) {
  pair<mi, mi> ret;
  ret.first = 0;
  ret.second = 0;
  vector<pair<mi, mi>> vec;
  for (auto adj : t[v]) {
    if (adj == p) continue;
    auto tmp = dfs(adj, v);
    vec.push_back(tmp);
  }
  if (vec.empty()) {
    if (x[v] == 1)
      return {1, 0};
    else
      return {0, 1};
  }
  if (x[v] == 1) {
    ret.first = 1;
    for (auto elm : vec) {
      ret.first *= (elm.first + elm.second);
    }
    return ret;
  } else {
    ret.first = 0;
    ret.second = 1;
    mi sum = 1;
    for (auto elm : vec) {
      ret.second *= (elm.first + elm.second);
      sum *= elm.first + elm.second;
    }
    for (auto elm : vec) {
      ret.first += (sum / (elm.first + elm.second)) * elm.first;
    }
    return ret;
  }
  return ret;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n;
  t.resize(n);
  x.resize(n);
  for (ll i = (0); i < (ll)(n - 1); i++) {
    int p;
    cin >> p;
    t[i + 1].push_back(p);
    t[p].push_back(i + 1);
  }
  for (ll i = (0); i < (ll)(n); i++) cin >> x[i];
  auto ans = dfs(0, -1);
  cout << ans.first << endl;
  return 0;
}
