CF1230E Kamil and Making a Stream

题目大意是求
\(\sum_{v,fa,lca(v,fa)=fa}gcd(v \to fa)\)

容易发现 \(\gcd\) 只会变小，所以根据这玩意是从上到下的，每次暴力一下就可以了，\(\gcd\)数量不会超过\(\log\)，所以复杂度大概是 \(n \log n\)

// powered by c++11
// by Isaunoya
#include 
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
using namespace std;
using db = double;
using ll = long long;
using uint = unsigned int;
#define int long long
using pii = pair;
#define ve vector
#define Tp template
#define all(v) v.begin(), v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
#define fir first
#define sec second
// the cmin && cmax
Tp void cmax(T& x, const T& y) {
  if (x < y) x = y;
}
Tp void cmin(T& x, const T& y) {
  if (x > y) x = y;
}
// sort , unique , reverse
Tp void sort(ve& v) { sort(all(v)); }
Tp void unique(ve& v) {
  sort(all(v));
  v.erase(unique(all(v)), v.end());
}
Tp void reverse(ve& v) { reverse(all(v)); }
const int SZ = 0x191981;
struct FILEIN {
  ~FILEIN() {}
  char qwq[SZ], *S = qwq, *T = qwq, ch;
  char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }
  FILEIN& operator>>(char& c) {
    while (isspace(c = GETC()))
      ;
    return *this;
  }
  FILEIN& operator>>(string& s) {
    while (isspace(ch = GETC()))
      ;
    s = ch;
    while (!isspace(ch = GETC())) s += ch;
    return *this;
  }
  Tp void read(T& x) {
    bool sign = 1;
    while ((ch = GETC()) < 0x30)
      if (ch == 0x2d) sign = 0;
    x = (ch ^ 0x30);
    while ((ch = GETC()) > 0x2f) x = x * 0xa + (ch ^ 0x30);
    x = sign ? x : -x;
  }
  FILEIN& operator>>(int& x) { return read(x), *this; }
  FILEIN& operator>>(signed& x) { return read(x), *this; }
  FILEIN& operator>>(unsigned& x) { return read(x), *this; }
} in;
struct FILEOUT {
  const static int LIMIT = 0x114514;
  char quq[SZ], ST[0x114];
  signed sz, O;
  ~FILEOUT() { flush(); }
  void flush() {
    fwrite(quq, 1, O, stdout);
    fflush(stdout);
    O = 0;
  }
  FILEOUT& operator<<(char c) { return quq[O++] = c, *this; }
  FILEOUT& operator<<(string str) {
    if (O > LIMIT) flush();
    for (char c : str) quq[O++] = c;
    return *this;
  }
  Tp void write(T x) {
    if (O > LIMIT) flush();
    if (x < 0) {
      quq[O++] = 0x2d;
      x = -x;
    }
    do {
      ST[++sz] = x % 0xa ^ 0x30;
      x /= 0xa;
    } while (x);
    while (sz) quq[O++] = ST[sz--];
    return;
  }
  FILEOUT& operator<<(int x) { return write(x), *this; }
  FILEOUT& operator<<(signed x) { return write(x), *this; }
  FILEOUT& operator<<(unsigned x) { return write(x), *this; }
} out;

int n, ans;
const int maxn = 1e5 + 10;
vector g[maxn];
int val[maxn];
int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }
const int mod = 1e9 + 7;
void dfs(vector vc, int u, int p) {
  vector nvc;
  nvc.pb(val[u], 1ll);
  ans = (ans + val[u]) % mod;
  for (pii x : vc) {
    int y = x.first;
    int c = x.second;
    int ny = gcd(y, val[u]);
    ans = (ans + ny * c) % mod;
    if (ny == nvc.back().first)
      nvc.back().second += c;
    else
      nvc.pb(ny, c);
  }
  nvc.swap(vc);
  for (int v : g[u])
    if (v != p) dfs(vc, v, u);
}
signed main() {
#ifdef _WIN64
  freopen("testdata.in", "r", stdin);
#else
  ios_base ::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
#endif
  // code begin.
  in >> n;
  for (int i = 0; i < n; i++) {
    in >> val[i];
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    in >> u >> v;
    --v, --u;
    g[u].pb(v);
    g[v].pb(u);
  }
  dfs({}, 0, -1);
  out << ans << '\n';
  return 0;
  // code end.
}