#include<bits/stdc++.h>
using namespace std;
using i64 = long long;
#define int i64
#define vi vector<int>
#define vvi vector<vi>
#define vb vector<bool>
#define pii pair<int, int>
#define fi first
#define se second
#define sz(x) (int)(x).size()
inline void solve(){
int n, m;
cin >> n >> m;
vector<pii> a(m); // (x, y), x = a[i], y = t[i], each hand can be descriped by a line heading downwards and connecting points, such that no line exceeds an angle of 45 degrees from the vertical (in both directions).
for(int i = 0; i < m; i++){
cin >> a[i].se;
}
for(int i = 0; i < m; i++){
cin >> a[i].fi;
}
for(int i = 0; i < m; i++){ // rotate the grid 45 degrees clockwise. Now all lines must connect points where one point is the bottom-left of another. This grid can be viewed as a DAG.
a[i] = {a[i].fi - a[i].se, a[i].fi + a[i].se};
}
// The question now becomes: How do we decompose this DAG into the minimum number of chains such that each chain follows the edges on that DAG?
// By Dilworth's theorem the minimum partitioning of a DAG into chains is equal to the size of the maximum antichain.
// Maximum antichain -> Finding a line connecting the maximum number of points such that the line can only head bottom-right.
// Finding this line is the equivalent of sorting the points by their y (t) values and finding the longest increasing subsequence by their x (a) values
sort(a.begin(), a.end(), [&](const pii &lhs, const pii &rhs){return lhs.se < rhs.se;} );
auto find_lis = [&]() -> int {
vi lis;
for(int i = 0; i < m; i++){
int pos = lower_bound(lis.begin(), lis.end(), a[i].fi) - lis.begin();
if(pos == sz(lis)){
lis.emplace_back(a[i].fi);
}
else{
lis[pos] = a[i].fi;
}
}
return sz(lis);
};
cout << find_lis() << "\n";
return;
}
signed main(){
ios_base::sync_with_stdio(false); cin.tie(0);
int t = 1;
//cin >> t;
while(t--){
solve();
}
return 0;
}
