F: MOD Rush

problem

Given a positive integer sequence A of length N and a positive integer sequence B of length M.

For all (i, j) (1 \ leq i \ leq N, 1 \ leq j \ leq M), find the remainder of A_i divided by B_j and output the sum of them.

Input format


N M
A_1 A_2 ... A_N
B_1 B_2 ... B_M


Constraint

* 1 \ leq N, M \ leq 2 \ times 10 ^ 5
* 1 \ leq A_i, B_i \ leq 2 \ times 10 ^ 5
* All inputs are given as integers



Output format

Print the answer on one line. Please start a new line at the end.

Input example 1


3 3
5 1 6
2 3 4


Output example 1


9

* Consider the remainder of dividing the first element of sequence A by each element of sequence B. Dividing 5 by 2 gives too much 1, dividing by 3 gives too much 2, and dividing by 4 gives too much 1.
* Similarly, considering the second element, too much is 1, 1, 1, respectively.
* Considering the third element, too much is 0, 0, 2, respectively.
* If you add up too much, you get 1 + 2 + 1 + 1 + 1 + 1 + 0 + 0 + 2 = 9, so 9 is output.



Input example 2


twenty four
2 7
3 3 4 4


Output example 2


16

* The same value may be included more than once in the sequence, but the sum is calculated by calculating too much for each element.



Input example 3


3 1
12 15 21
3


Output example 3


0





Example

Input

3 3
5 1 6
2 3 4


Output

9
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// mmqHl(dk_.(_.-=~`.`.1-..._~-1.``_:`(??<_~(`.--.&_.`.<(;<...._.._<..`..._Xg@@@@@@
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*/

#include <bits/stdc++.h>
using namespace std;

using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;
using LL = long long;
using VL = vector<LL>;
using VVL = vector<VL>;
using PLL = pair<LL, LL>;
using VS = vector<string>;

#define ALL(a)  begin((a)),end((a))
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define MT make_tuple
#define SZ(a) int((a).size())
#define SORT(c) sort(ALL((c)))
#define RSORT(c) sort(RALL((c)))
#define UNIQ(c) (c).erase(unique(ALL((c))), end((c)))

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

#define FF first
#define SS second

template<class S, class T>
istream& operator>>(istream& is, pair<S,T>& p){
  return is >> p.FF >> p.SS;
}
template<class T>
istream& operator>>(istream& is, vector<T>& xs){
  for(auto& x: xs)
	is >> x;
  return is;
}
template<class S, class T>
ostream& operator<<(ostream& os, const pair<S,T>& p){
  return os << p.FF << " " << p.SS;
}
template<class T>
ostream& operator<<(ostream& os, const vector<T>& xs){
  for(unsigned int i=0;i<xs.size();++i)
	os << (i?" ":"") << xs[i];
  return os;
}
template<class T>
void maxi(T& x, T y){
  if(x < y) x = y;
}
template<class T>
void mini(T& x, T y){
  if(x > y) x = y;
}
void debug(istringstream&){}
template <char sep=',', class Head, class... Tail>
void debug(istringstream& iss, Head&& head, Tail&&... tail)
{
  string name;
  getline(iss, name, ',');
  cout << sep << name << "=" << head;
  debug(iss, forward<Tail>(tail)...);
}

#ifdef PYONPOI
#define DEBUG(...) \
  do{ \
    istringstream ss(#__VA_ARGS__); \
    debug<' '>(ss, __VA_ARGS__); \
    cout<<endl; \
  }while(0)
#else
#define DEBUG
#endif


const double EPS = 1e-10;
const double PI  = acos(-1.0);
const LL MOD = 1e9+7;

int main(){
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  int N, M;
  cin >> N >> M;
  VI as(N), bs(M);
  cin >> as >> bs;

  SORT(bs);
  const int MAX = 2e5 + 10;
  VL acc(MAX);
  for(auto a : as) acc[a+1]++;
  FOR(i,1,MAX) acc[i] += acc[i-1];

  LL ans = 0;
  map<LL,LL> mp;
  REP(j,M){
    LL b = bs[j];
    if(mp.count(b)){
      ans += mp[b];
      continue;
    }
    LL sum = 0;
    int ub = (MAX + b - 1) / b;
    for(int q=1;q<=ub;++q){
      int beg = min((LL)MAX-1, q * b);
      int ed = min((LL)MAX-1, q * b + b);
      sum += (acc[ed] - acc[beg]) * q;
    }
    ans += sum * b;
    mp[b] = sum * b;
  }

  LL sum = accumulate(ALL(as), 0ll) * M;
  cout << sum - ans << endl;

  return 0;
}

