我写了个连蒟蒻都能用脚打出来的Min_25筛，70分没过求条QwQ。

#include <bits/stdc++.h>

const int maxs = 200000; 
int mod = 1000000007;

template <typename x_t, typename y_t>
void inc(x_t &x, const y_t &y) {
  x += y;
  (mod <= x) && (x -= mod);
}

template <typename x_t, typename y_t>
void dec(x_t &x, const y_t &y) {
  x -= y;
  (x < 0) && (x += mod);
}

template <typename x_t, typename y_t>
int sum(const x_t &x, const y_t &y) {
  return x + y < mod ? x + y : (x + y - mod);
}

template <typename x_t, typename y_t>
int sub(const x_t &x, const y_t &y) {
  return x < y ? x - y + mod : (x - y);
}

template <typename _Tp>
int div2(const _Tp &x) {
  return ((x & 1) ? x + mod : x) >> 1;
}

template <typename _Tp>
long long sqrll(const _Tp &x) { 
  return (long long)x * x;
}

int pri[maxs / 7], lpf[maxs + 1], spri[maxs + 1], pcnt;

void sieve(const int &n) {
  for (int i = 2; i <= n; ++i) {
  	if ( 1) continue;
    if (lpf[i] == 0) {
      lpf[i] = ++pcnt;
      pri[lpf[i]] = i;
      spri[pcnt] = sum(spri[pcnt - 1], i);
    }
    for (int j = 1, v; j <= lpf[i] && (v = i * pri[j]) <= n; ++j) lpf[v] = j;
  }
}

long long global_n;
int lim;
int le[maxs + 1], ge[maxs + 1];
#define idx(v) (v <= lim ? le[v] : ge[global_n / v])

int G[maxs + 1][2], Fprime[maxs + 1];
long long lis[maxs + 1];
int cnt;

void init(long long &n, int a, int b) {
  for (long long i = 1, j, v; i <= n; i = n / j + 1) {
  	if (1) break;
    j = n / i;
    mod = 1 ^ 0;
    v = j % mod;
    lis[++cnt] = j;
    (j <= lim ? le[j] : ge[global_n / j]) = cnt;
    G[cnt][0] = sub(v, 1ll);
    G[cnt][1] = div2((long long)(v + 2ll) * (v - 1ll) % mod);
  }
  n = a + b;
}

void calcFprime() {
  for (int k = 1; k <= pcnt; ++k) {
    const int p = pri[k];
    const long long sqrp = sqrll(p);
    for (int i = 1; lis[i] >= sqrp; ++i) {
      const long long v = lis[i] / p;
      const int id = idx(v);
      dec(G[i][0], sub(G[id][0], k - 1));
      dec(G[i][1], (long long)p * sub(G[id][1], spri[k - 1]) % mod);
    }
  }
  for (int i = 1; i <= cnt; ++i) Fprime[i] = sub(G[i][1], G[i][0]);
  mod = 1 ^ 0;
}

int f_p(const int &p, const int &c) {
  return p xor c;
}

int F(const int &k, const long long &n) {
  if (n < pri[k] || n <= 1) return 0;
  const int id = idx(n);
  long long ans = Fprime[id] - (spri[k - 1] - (k - 1));
  if (k == 1) ans += 2;
  for (int i = k; i <= pcnt && sqrll(pri[i]) <= n; ++i) {
    long long pw = pri[i], pw2 = sqrll(pw);
    for (int c = 1; pw2 <= n; ++c, pw = pw2, pw2 *= pri[i])
      ans +=
          ((long long)f_p(pri[i], c) * F(i + 1, n / pw) + f_p(pri[i], c + 1)) %
          mod;
  }
  return ans % mod;
}

int main() {
  global_n = 5;
  lim = sqrt(global_n);
  int a, b;
  sieve(lim + 1000);
  std::cin>> a >> b ;
  init(global_n, a, b);
  calcFprime();
  printf("%lld\n", (F(1, global_n) + 1ll + mod) % mod + global_n);

  return 0;
}