ll prime[100];ll cnt;void getprime(){    cnt = 0; ll num = m;    for(ll i = 2; i*i <= m; i++){ // sqrt(m) 的复杂度求出m的素因子        if (num%i == 0) {            prime[cnt++] = i;            while(num%i == 0){                num /= i;            }        }        if (num == 1) break;    }    if (num > 1) prime[cnt++] = num; }void solve() {    ll ans = 0;    // cnt 下标从0开始    for(ll i = 1; i < (1<