description: Mr. Chanek has an array a of n integers. The prettiness value of a is denoted as:
  
  $$$∑_{i=1}^{n} {∑_{j=1}^{n} {\gcd(a_i, a_j) ⋅ \gcd(i, j)}}$$$
  
  where \gcd(x, y) denotes the greatest common divisor (GCD) of integers x and y.
  
  In other words, the prettiness value of an array a is the total sum of \gcd(a_i, a_j) ⋅ \gcd(i, j) for all pairs (i, j).
  
  Help Mr. Chanek find the prettiness value of a, and output the result modulo 10^9 + 7!
  
  Input
  
  The first line contains an integer n (2 ≤ n ≤ 10^5).
  
  The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^5).
  
  Output
  
  Output an integer denoting the prettiness value of a modulo 10^9 + 7.
  
  Example
  
  Input
  
  
  5
  3 6 2 1 4
  
  
  Output
  
  
  77
solution: #include <bits/stdc++.h>
  using namespace std;
  const int MOD = (int)1e9 + 7;
  const int FFTMOD = 119 << 23 | 1;
  const int INF = (int)1e9 + 23111992;
  const long long LINF = (long long)1e18 + 23111992;
  const long double PI = acos((long double)-1);
  const long double EPS = 1e-9;
  inline long long gcd(long long a, long long b) {
    long long r;
    while (b) {
      r = a % b;
      a = b;
      b = r;
    }
    return a;
  }
  inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  inline long long fpow(long long n, long long k, int p = MOD) {
    long long r = 1;
    for (; k; k >>= 1) {
      if (k & 1) r = r * n % p;
      n = n * n % p;
    }
    return r;
  }
  template <class T>
  inline int chkmin(T& a, const T& val) {
    return val < a ? a = val, 1 : 0;
  }
  template <class T>
  inline int chkmax(T& a, const T& val) {
    return a < val ? a = val, 1 : 0;
  }
  inline unsigned long long isqrt(unsigned long long k) {
    unsigned long long r = sqrt(k) + 1;
    while (r * r > k) r--;
    return r;
  }
  inline long long icbrt(long long k) {
    long long r = cbrt(k) + 1;
    while (r * r * r > k) r--;
    return r;
  }
  inline void addmod(int& a, int val, int p = MOD) {
    if ((a = (a + val)) >= p) a -= p;
  }
  inline void submod(int& a, int val, int p = MOD) {
    if ((a = (a - val)) < 0) a += p;
  }
  inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
  inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
  inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
  inline int sign(long double x, long double y) { return sign(x - y); }
  mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
  inline int mrand() { return abs((int)mt()); }
  inline int mrand(int k) { return abs((int)mt()) % k; }
  void chemthan() {
    int C = 1e5 + 5;
    vector<int> sp(C);
    for (int i = (1); i < (C); ++i) sp[i] = i;
    for (int i = (2); i < (C); ++i)
      if (sp[i] == i) {
        for (int j = i + i; j < C; j += i) {
          chkmin(sp[j], i);
        }
      }
    vector<int> mob(C, 1);
    for (int i = (2); i < (C); ++i) {
      int t = i;
      int pv = -1;
      while (1 < t) {
        if (pv == sp[t]) {
          mob[i] = 0;
        }
        mob[i] *= -1;
        pv = sp[t];
        t /= sp[t];
      }
    }
    vector<vector<int> > dvs(C);
    vector<vector<int> > dvs2(C);
    for (int i = (1); i < (C); ++i) {
      for (int j = i; j < C; j += i) {
        dvs[j].push_back(i);
        if (mob[j / i]) dvs2[j].push_back(i);
      }
    }
    int n;
    cin >> n;
    vector<int> a(n);
    vector<vector<int> > g(C);
    for (int i = (0); i < (n); ++i) {
      cin >> a[i];
      for (int d : dvs[i + 1]) {
        g[d].push_back(a[i]);
      }
    }
    vector<int> c(C);
    vector<int> f(C);
    vector<int> x(C);
    vector<int> y(C);
    int res = 0;
    for (int i = (1); i < (C); ++i)
      if (int((g[i]).size())) {
        vector<int> tmp2;
        vector<int> tmp;
        for (int u : g[i]) {
          c[u] = 0;
        }
        for (int u : g[i]) {
          if (!c[u]++) {
            tmp2.push_back(u);
          }
        }
        for (int u : tmp2) {
          for (int d : dvs[u]) {
            if (!f[d]) {
              tmp.push_back(d);
            }
            f[d] += c[u];
          }
        }
        for (int u : tmp) {
          f[u] = mult(f[u], f[u]);
        }
        int sum = 0;
        for (int u : tmp) {
          for (int v : dvs2[u]) {
            if (mob[u / v]) {
              if (mob[u / v] == 1) {
                addmod(x[v], f[u]);
              } else {
                submod(x[v], f[u]);
              }
            }
          }
        }
        for (int u : tmp) {
          addmod(sum, mult(x[u], u));
        }
        y[i] = sum;
        for (int u : tmp) f[u] = x[u] = 0;
      }
    for (int i = (C)-1; i >= (1); --i) {
      for (int j = i + i; j < C; j += i) {
        submod(y[i], y[j]);
      }
      addmod(res, mult(i, y[i]));
    }
    cout << res << "\n";
  }
  int32_t main(int32_t argc, char* argv[]) {
    ios_base::sync_with_stdio(0), cin.tie(0);
    if (argc > 1) {
      assert(freopen(argv[1], "r", stdin));
    }
    if (argc > 2) {
      assert(freopen(argv[2], "wb", stdout));
    }
    chemthan();
    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
  }
  