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题解讲的很清楚了，直接看代码就懂了

题解:http://bestcoder.hdu.edu.cn/blog/2018-multi-university-training-contest-6-solutions-by-%e7%a6%8f%e5%b7%9e%e5%a4%a7%e5%ad%a6/

#include<stdio.h>
#include<iostream>
#include<vector>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<cmath>
#include<set>
#include<map>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define all(a) a.begin(),a.end()
using namespace std;
typedef vector<int> vi;
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;

const ll mod = 1e9+7;
const int N = 2e6+6;
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
bool ls(const db &a, const db &b) { return a + eps < b; }
bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); };
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
ll inv[N], f[N], fac[N];
ll a[N];
void init() {
    f[0] = 1; f[1] = 2;
    fac[0] = fac[1] = 1;
    inv[1]=1;
    rep(i, 2, N) {
        fac[i] = fac[i-1] * (ll)i % mod;
        inv[i] = kpow(fac[i], mod-2);
        f[i] = (f[i-1] * f[i-2]) % mod;
    }
}
ll C(int n, int m) {
    if(n < m) return 0ll;
    if(m==0 || n==m) return 1ll;
    if(n-1==m || m==1) return n;
    return fac[n] * inv[m] % mod * inv[n-m] % mod;
}
int main(){
    init();
    int T;
    scanf("%d",&T);
    while(T--) {
        int n, m;
        scanf("%d%d",&n,&m);
        mt(a, 0);
        ll sum = C(n+m-1, n), ans = 0;
        repd(g,n,1) {
            if(n%g==0) {
                ll j = n / g;
                a[g] = C(j+m-1,j) % mod;
                repp(k,g+g,n + 1,g) a[g] = (a[g] - a[k] + mod) % mod;
                ans = ans + a[g] * (f[g] - 1 + mod) % mod; ans %= mod;
            }
        }
        ans = ans * kpow(sum, mod-2) % mod;
        printf("%lld
",ans);
    }
    return 0;
}