来自"NOCOW"

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Solution From Xcheng:

计算质因数个数的快速方法：

质因数K在1~N中出现在次数为

F(N)=sigma(n div (k^i) )

所以计算出F(N)-F(N-M)-F(M) 再统计一遍就行了。

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把C(n,m)展开得C=N!/(M!·(N−M)!),再分解质因数即可.

program cao;
var
  data,p:array[0..50000] of longint;
  a,b,c,d,e,f,g,h,i,j,k,l,n,m,q,total,ans:longint;
 
function prime(x:longint):boolean;
var
  i:longint;
begin
  if x=2 then exit(true);
  for i:=2 to trunc(sqrt(x)) do
    if x mod i=0 then exit(false);
  exit(true);
end;
 
procedure work(a,b:longint);
var
  i,j,k:longint;
begin
  for i:=1 to total do
  begin
    if a=1 then exit;
    if data[i]>a then exit;
    while a mod data[i]=0 do
    begin
      p[i]:=p[i]+b;
      a:=a div data[i];
    end;
  end;
end;
 
procedure init;
begin
  read(n,m);
  if m<n-m then m:=n-m;
  total:=0;
  for i:=2 to n do
    if prime(i) then
    begin
      inc(total);
      data[total]:=i;
    end;
end;
 
procedure main;
begin
  for i:=m+1 to n do
    work(i,1);
  for i:=2 to n-m do
    work(i,-1);
end;
 
procedure print;
begin
  ans:=0;
  for i:=1 to total do
    if p[i]>0 then inc(ans);
  writeln(ans);
end;
 
begin
  init;
  main;
  print;
end.