#include <bits/stdc++.h>
using namespace std;
typedef long double ld;

const ld PI=acos(-1.0);
typedef complex<ld> Complex;
inline int rev_incr(int a, int n) { int msk = n/2, cnt=0;
  while ( a&msk ) { cnt++; a<<=1; }
  a &= msk-1; a |= msk;
  while ( cnt-- ) a >>= 1;
  return a; }
vector<Complex> FFT(vector<Complex> v, int dir=1) {
  Complex wm,w,u,t; int n = v.size(); vector<Complex> V(n);
  for (int k=0,a=0; k<n; ++k, a=rev_incr(a,n))
    V[a] = v[k] / ld(dir>0 ? 1 : n);
  for (int m=2; m<=n; m<<=1) {
    wm = polar( (ld)1, dir*2*PI/m );
    for (int k=0; k<n; k+=m) {
      w = 1;
      for (int j=0; j<m/2; ++j, w*=wm) {
        u = V[k + j];
        t = w * V[k + j + m/2];
        V[k + j] = u + t;
        V[k + j + m/2] = u - t;
      } } } return V; }
//! See problem 30-6 in CLRS for more details on the integer FFT
//! hints for integer version:
//! 440564289 and 1713844692 are inverses and 2^27th roots of 1 mod p=(15<<27)+1
//! Precompute inverses for integer FFT, otherwise it will be very slow.
// Multiply two polynomials sum_i a[i]x^i and sum_i b[i]x^i
vector<ld> convolution(vector<ld> a, vector<ld> b) {
  int sz = a.size()+b.size()-1;
  int n  = 1<<int(ceil(log2(sz)));
  vector<Complex> av(n,0), bv(n,0), cv;
  for (int i=0; i<a.size(); i++) av[i] = a[i];
  for (int i=0; i<b.size(); i++) bv[i] = b[i];
  cv = FFT(bv); bv = FFT(av);
  for (int i=0; i<n; i++) av[i] = bv[i]*cv[i];
  cv = FFT(av, -1);
  // cv is now the convolution of a and b: cv[k] = sum_(i+j=k) a[i]*b[j]
  vector<ld> c(sz);
  for (int i=0; i<sz; i++)
    c[i] = cv[i].real(); // if result should be int, use int(cv[i].real()+0.5)
  return c; }

int main(){
    ios::sync_with_stdio(0);
    int n; cin>>n;
    vector<ld> a;
    for (int i=0;i<n;i++){
        int x; cin>>x;
        while (a.size()<=x) a.push_back(0);
        a[x]=1;
    }
    int ans=0;
    int m; cin>>m;
    vector<ld> res=convolution(a,a);
    //for (ld x:a) cout<<x<<" "; cout<<endl;
    //for (ld x:res) cout<<x<<" "; cout<<endl;
    while (m--){
        int x; cin>>x;
        if (((x<res.size())&&(res[x]>0.5))||((x<a.size())&&(a[x]>0.5))) {
            ans++;
        }
    }
    cout<<ans<<endl;
    return 0;
}

//g++ -std=c++11 -O2 -Wfatal-errors -Im -Wall -Wextra -o "output.txt" "code.cpp"