Kamil likes streaming the competitive programming videos. His MeTube channel has recently reached 100 million subscribers. In order to celebrate this, he posted a video with an interesting problem he couldn't solve yet. Can you help him?

You're given a tree — a connected undirected graph consisting of n vertices connected by n - 1 edges. The tree is rooted at vertex 1. A vertex u is called an ancestor of v if it lies on the shortest path between the root and v. In particular, a vertex is an ancestor of itself.

Each vertex v is assigned its beauty x_v — a non-negative integer not larger than 10^{12}. This allows us to define the beauty of a path. Let u be an ancestor of v. Then we define the beauty f(u, v) as the greatest common divisor of the beauties of all vertices on the shortest path between u and v. Formally, if u=t_1, t_2, t_3, ..., t_k=v are the vertices on the shortest path between u and v, then f(u, v) = \gcd(x_{t_1}, x_{t_2}, ..., x_{t_k}). Here, \gcd denotes the greatest common divisor of a set of numbers. In particular, f(u, u) = \gcd(x_u) = x_u.

Your task is to find the sum

$$$ ∑_{u is an ancestor of v} f(u, v). $$$

As the result might be too large, please output it modulo 10^9 + 7.

Note that for each y, \gcd(0, y) = \gcd(y, 0) = y. In particular, \gcd(0, 0) = 0.

Input

The first line contains a single integer n (2 ≤ n ≤ 100 000) — the number of vertices in the tree.

The following line contains n integers x_1, x_2, ..., x_n (0 ≤ x_i ≤ 10^{12}). The value x_v denotes the beauty of vertex v.

The following n - 1 lines describe the edges of the tree. Each of them contains two integers a, b (1 ≤ a, b ≤ n, a ≠ b) — the vertices connected by a single edge.

Output

Output the sum of the beauties on all paths (u, v) such that u is ancestor of v. This sum should be printed modulo 10^9 + 7.

Examples

Input


5
4 5 6 0 8
1 2
1 3
1 4
4 5


Output


42


Input


7
0 2 3 0 0 0 0
1 2
1 3
2 4
2 5
3 6
3 7


Output


30

Note

The following figure shows all 10 possible paths for which one endpoint is an ancestor of another endpoint. The sum of beauties of all these paths is equal to 42:

<image>
#include <bits/stdc++.h>
using namespace std;
using namespace chrono;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long int advceil(long long int num, long long int den) {
  return (num % den == 0 ? num / den : num / den + 1);
}
long long int lstbt(long long int val) {
  long long int msk = val & (val - 1);
  return log2(val ^ msk);
}
long long int modmul(long long int a, long long int b) {
  if (a < 1e9 and b < 1e9) return (a * b) % 1000000007;
  a %= 1000000007;
  b %= 1000000007;
  long long int res = 0;
  while (b > 0) {
    if (b & 1) res = (res + a) % 1000000007;
    a = (a * 2) % 1000000007;
    b /= 2;
  }
  return res;
}
long long int modexpo(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = modmul(res, a);
    a = modmul(a, a);
    b /= 2;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  return a == 0 ? b : gcd(b % a, a);
}
vector<long long int> CALCfactor(long long int n) {
  vector<long long int> factor(n + 2, 0);
  for (long long int i = 4; i <= n; i += 2) factor[i] = 2;
  for (long long int j = 3; j <= n; j += 2) {
    if (factor[j]) continue;
    for (long long int i = 2 * j; i <= n; i += j) factor[i] = j;
  }
  return factor;
}
vector<long long int> CALCprimeNUM(long long int n) {
  vector<long long int> factor = CALCfactor(n);
  vector<long long int> primeNUM;
  primeNUM.reserve(n + 2);
  for (long long int i = 2; i <= n; i++)
    if (!factor[i]) primeNUM.push_back(i);
  return primeNUM;
}
vector<long long int> CALCprimeFACTOR(long long int n) {
  vector<long long int> factor = CALCfactor(n);
  vector<long long int> ans;
  while (factor[n] != 0) {
    ans.push_back(factor[n]);
    n /= factor[n];
  }
  if (n > 1) ans.push_back(n);
  return ans;
}
vector<long long int> unique(vector<long long int> x) {
  sort(x.begin(), x.end());
  set<long long int> s;
  vector<long long int> ans;
  ans.reserve(x.size());
  for (auto elem : x) s.insert(elem);
  for (auto elem : s) ans.push_back(elem);
  return ans;
}
pair<vector<long long int>, vector<long long int> > getFact(long long int n) {
  vector<long long int> fact(n + 1, 1), invfact(n + 1, 1);
  for (long long int i = 1; i <= n; i++)
    fact[i] = (i * (fact[i - 1])) % 1000000007;
  for (long long int i = 1; i <= n; i++)
    invfact[i] = (modexpo(i, 1000000007 - 2) * invfact[i - 1]) % 1000000007;
  return {fact, invfact};
}
void compress(vector<long long int>& arr, long long int n) {
  map<long long int, vector<long long int> > pos;
  for (long long int i = 1; i <= n; i++) pos[arr[i]].push_back(i);
  long long int cnt = 1;
  for (auto itr : pos) {
    for (auto elem : itr.second) arr[elem] = cnt;
    cnt++;
  }
}
bool rcomp(long long int a, long long int b) { return a > b; }
long long int beauty[100001] = {};
vector<long long int> edges[100001] = {};
long long int ans = 0;
void dfs(long long int src, long long int par,
         map<long long int, long long int>& freq) {
  map<long long int, long long int> resources;
  for (auto elem : freq) {
    resources[gcd(elem.first, beauty[src])] += elem.second;
  }
  resources[beauty[src]]++;
  for (auto elem : resources) {
    ans += (elem.first * elem.second);
    ans %= 1000000007;
  }
  for (auto elem : edges[src]) {
    if (elem == par) continue;
    dfs(elem, src, resources);
  }
}
void solve() {
  long long int n;
  cin >> n;
  long long int sum = 0;
  for (long long int i = 1; i <= n; i++) cin >> beauty[i];
  for (long long int i = 1; i <= n - 1; i++) {
    long long int x, y;
    cin >> x >> y;
    edges[x].push_back(y);
    edges[y].push_back(x);
  }
  map<long long int, long long int> freq;
  dfs(1, 0, freq);
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  auto start1 = high_resolution_clock::now();
  solve();
  auto stop1 = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop1 - start1);
}
