[cf920G][容斥原理+二分]

https://codeforc.es/contest/920/problem/G

G. List Of Integers

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's denote as L(x, p) an infinite sequence of integers y such that gcd(p, y) = 1 and y > x (where gcd is the greatest common divisor of two integer numbers), sorted in ascending order. The elements of L(x, p) are 1-indexed; for example, 9, 13 and 15 are the first, the second and the third elements of L(7, 22), respectively.

You have to process t queries. Each query is denoted by three integers x, p and k, and the answer to this query is k-th element of L(x, p).

Input

The first line contains one integer t (1 ≤ t ≤ 30000) — the number of queries to process.

Then t lines follow. i-th line contains three integers x, p and k for i-th query (1 ≤ x, p, k ≤ 106).

Output

Print t integers, where i-th integer is the answer to i-th query.

Examples

input

Copy

3
7 22 1
7 22 2
7 22 3

output

Copy

9
13
15

input

Copy

5
42 42 42
43 43 43
44 44 44
45 45 45
46 46 46

output

Copy

187
87
139
128
141
题意：求出与p互质并且>x的第k小数
题解：首先求出p所有的质因子，状压枚举出质因子相乘的情况，比如000001表示只有一个最小的质因子，则在1-y中有y/prime[1]个，再通过容斥即可求出与p不互质的个数，此时二分答案y即可

 #include<bits/stdc++.h>
 using namespace std;
 #define debug(x) cout<<#x<<" is "<<x<<endl;
 typedef long long ll;
 ll sol(ll x,vector<ll>&f){
     ll res=;
     int n=(int)f.size();
     for(int i=;i<(<<n);i++)
     {
         ll now=,sgn=-;
         for(int j=;j<n;j++)
             if(i&(<<j))now*=f[j],sgn*=-;
         res+=sgn*(x/now);
     }
     return x-res;
 }
 int main(){
     int t;
     scanf("%d",&t);
     while(t--){
         ll x,p,k;
         scanf("%lld%lld%lld",&x,&p,&k);
         vector<ll>g;
         for(ll i=;i*i<=p;i++){
             if(p%i==){
                 g.push_back(i);
                 while(p%i==){
                     p/=i;
                 }
             }
         }
         if(p>)g.push_back(p);
         ll r=1000000000000LL;
         ll l=;
         k+=sol(x,g);
         ll ans;
         while(l<=r){
             ll mid=(l+r)/;
             if(sol(mid,g)>=k){
                 ans=mid;
                 r=mid-;
             }
             else{
                 l=mid+;
             }
         }
         printf("%lld\n",ans);
     }
     return ;
 }