Problem: There are n cities in the Kingdom of Autumn, numbered from 1 to n. People can travel between any two cities using n-1 two-directional roads.

This year, the government decides to separate the kingdom. There will be regions of different levels. The whole kingdom will be the region of level 1. Each region of i-th level should be separated into several (at least two) regions of i+1-th level, unless i-th level is the last level. Each city should belong to exactly one region of each level and for any two cities in the same region, it should be possible to travel between them passing the cities in the same region only.

According to research, for each city i, there is a value a_i, which describes the importance of this city. All regions of the same level should have an equal sum of city importances.

Your task is to find how many plans there are to determine the separation of the regions that all the conditions are satisfied. Two plans are considered different if and only if their numbers of levels are different or there exist two cities in the same region of one level in one plan but in different regions of this level in the other plan. Since the answer may be very large, output it modulo 10^9+7.

Input

The first line contains one integer n (1 ≤ n ≤ 10^6) — the number of the cities.

The second line contains n integers, the i-th of which is a_i (1 ≤ a_i ≤ 10^9) — the value of each city.

The third line contains n-1 integers, p_1, p_2, …, p_{n-1}; p_i (p_i ≤ i) describes a road between cities p_i and i+1.

Output

Print one integer — the number of different plans modulo 10^9+7.

Examples

Input

4
1 1 1 1
1 2 3


Output

4

Input

4
1 1 1 1
1 2 2


Output

2

Input

4
1 2 1 2
1 1 3


Output

3

Note

For the first example, there are 4 different plans:

Plan 1: Level-1: \{1,2,3,4\}.

Plan 2: Level-1: \{1,2,3,4\}, Level-2: \{1,2\},\{3,4\}.

Plan 3: Level-1: \{1,2,3,4\}, Level-2: \{1\},\{2\},\{3\},\{4\}.

Plan 4: Level-1: \{1,2,3,4\}, Level-2: \{1,2\},\{3,4\}, Level-3: \{1\},\{2\},\{3\},\{4\}.

For the second example, there are 2 different plans:

Plan 1: Level-1: \{1,2,3,4\}.

Plan 2: Level-1: \{1,2,3,4\}, Level-2: \{1\},\{2\},\{3\},\{4\}.

For the third example, there are 3 different plans:

Plan 1: Level-1: \{1,2,3,4\}.

Plan 2: Level-1: \{1,2,3,4\}, Level-2: \{1,2\},\{3,4\}.

Plan 3: Level-1: \{1,2,3,4\}, Level-2: \{1,3\},\{2\},\{4\}.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
template <int MOD>
struct Integral {
  int v_ = 0;
  template <typename T>
  Integral(T v) : v_(norm(v)) {
    static_assert(std::is_integral<T>::value, "input should be an integral.");
  }
  Integral() = default;
  ~Integral() = default;
  template <typename T>
  T norm(T v) const {
    if (v >= MOD) v -= MOD;
    if (v < 0) v += MOD;
    if (v >= MOD || v < 0) v = (v % MOD + MOD) % MOD;
    return v;
  }
  int val() const { return v_; }
  Integral operator+(const Integral& rhs) const {
    return Integral(val() + rhs.val());
  }
  Integral operator-(const Integral& rhs) const {
    return Integral(val() - rhs.val());
  }
  Integral operator*(const Integral& rhs) const {
    return Integral(val() * 1LL * rhs.val());
  }
  Integral operator/(const Integral& rhs) const { return *this * rhs.inv(); }
  Integral& operator+=(const Integral& rhs) { return *this = *this + rhs; }
  Integral& operator-=(const Integral& rhs) { return *this = *this - rhs; }
  Integral& operator*=(const Integral& rhs) { return *this = *this * rhs; }
  Integral& operator/=(const Integral& rhs) { return *this = *this / rhs; }
  bool operator==(const Integral& rhs) const { return val() == rhs.val(); }
  bool operator!=(const Integral& rhs) const { return !(*this == rhs); }
  const Integral operator-() const { return Integral(-val()); }
  const Integral operator++() {
    v_ = norm(v_ + 1);
    return *this;
  }
  const Integral operator++(int) {
    Integral ret = *this;
    ++(*this);
    return ret;
  }
  const Integral operator--() {
    v_ = norm(v_ - 1);
    return *this;
  }
  const Integral operator--(int) {
    Integral ret = *this;
    --(*this);
    return ret;
  }
  Integral power(long long b) const {
    long long ret = 1 % MOD, a = v_;
    for (; b; b >>= 1, a = a * a % MOD)
      if (b & 1) ret = ret * a % MOD;
    return ret;
  }
  Integral inv() const { return power(MOD - 2); }
};
template <int MOD>
std::string to_string(const Integral<MOD>& v) {
  return std::string("Int<>{") + std::to_string(v.val()) + "}";
}
template <int MOD, bool kAllowBruteForce = false>
struct Binomial {
  std::vector<Integral<MOD>> factor, inv_factor;
  explicit Binomial(int n = 0) : factor(n + 1), inv_factor(n + 1) {
    factor[0] = 1;
    for (int i = 1; i <= n; ++i) factor[i] = factor[i - 1] * i;
    inv_factor[n] = factor[n].inv();
    for (int i = n; i >= 1; --i) inv_factor[i - 1] = inv_factor[i] * i;
  }
  ~Binomial() = default;
  template <typename T>
  Integral<MOD> operator()(T a, T b) const {
    if (a < b || b < 0) return 0;
    if (a < factor.size()) return factor[a] * inv_factor[b] * inv_factor[a - b];
  }
};
template <int MOD>
struct PowerTable : public std::vector<Integral<MOD>> {
  PowerTable(int n, const Integral<MOD>& g) {
    static_assert(sizeof(PowerTable) == sizeof(std::vector<Integral<MOD>>), "");
    this->resize(n + 1);
    this->at(0) = 1;
    this->at(1) = g;
    for (int i = 2; i < this->size(); ++i)
      this->at(i) = this->at(i - 1) * this->at(1);
  }
};
const int MOD = 1e9 + 7;
using Mint = Integral<MOD>;
using Binom = Binomial<MOD>;
using LL = long long;
LL gcd(LL a, LL b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::istream& reader = std::cin;
  int n;
  reader >> n;
  std::vector<LL> a(n);
  std::vector<LL> s(n);
  for (int i = 0; i < n; ++i) {
    reader >> a[i];
    s[i] = a[i];
  }
  1145141919810;
  std::vector<int> p(n, -1);
  for (int i = 1; i < n; ++i) {
    reader >> p[i];
    --p[i];
  }
  1145141919810;
  for (int i = n - 1; i > 0; --i) {
    s[p[i]] += s[i];
  }
  1145141919810;
  std::vector<int> cnt(n + 1);
  for (int i = 0; i < n; ++i) {
    LL x = s[0] / gcd(s[0], s[i]);
    if (x <= n) {
      cnt[x]++;
    }
  }
  for (int i = n; i >= 1; --i) {
    for (int j = i + i; j <= n; j += i) cnt[j] += cnt[i];
  }
  1145141919810;
  std::vector<Mint> way(n + 1);
  for (int i = n; i >= 1; --i) {
    if (cnt[i] == i) {
      way[i] = 1;
      for (int j = i + i; j <= n; j += i) way[i] += way[j];
    }
  }
  1145141919810;
  printf("%d\n", way[1].val());
}